By how much does the gravitational potential energy of a 55- kg pole vaulter change if her center of mass rises about 4.0 m during the jump?
Express your answer to two significant figures and include the appropriate units.

Answers

Answer 1

The change in the gravitational potential energy of the pole vaulter is 2,156 J.

Change in the gravitational potential energy

The change in the gravitational potential energy of the pole vaulter is calculated as follows;

ΔP.E = mg(Δh)

where;

m is massΔh is change in height

ΔP.E = (55)(9.8)(4)

ΔP.E = 2,156 J

Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.

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Related Questions

Define the term work and state its unit. An ant is dragging a house-fly and the elephant is pushing a big tree which is not moving. Who is doing work, the ant or the elephant? Justify your answer. 922.5 205​

Answers

Ant is performing a work

what is work?

Work is the force applied on an individual with respect to displacement.

Work = Force × displacement

Unit is Nm

Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.

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The ant works, but the elephant does not.

Who works, how do find the ant and the elephant?

Work done = Force × Displacement.

If there are ants and houseflies,

Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.

In the case of the elephant and the tree,

When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.

Work done = Force × Displacement

= Force × 0

= 0

Therefore,

The ant works, but the elephant does not.

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if f=30 m=4 calculate value of m in equation
m=f/a

Answers

Answer:

a = 7.5

Explanation:

m = f/a

f = 30

m = 4

Thus;

a = f/m

a = 30/4

a = 7.5

What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?

Answers

0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.

What is a resistor?

A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.

The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:

R = R0[1+ alpha(T-T0)]  where:

R is the new resistance we are looking for

alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x [tex]10^{-4}[/tex](1/°c)

T0 is the standard temperature =25.8°C

R0 is the resistance at T0 = 0.0200 ohms

T is the temperature at which we want to get R = 0

Substitute in the equation to get R as follows:

R = 0.0200 [1+( ₋ 4.8 x [tex]10^{-4}[/tex]) (0-25.8)] = 0.02020 ohm

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please help me with this physics question ASAP​

Answers

Answer:

See below

Explanation:

With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings

V = IR

I = V/R = 5 / (10+5+5) = .25 A

b) With S1 closed   5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm

 then the series circuit current becomes  

     5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps

            ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3        .273 *   10 / 15 =.182 amps

I need help with my homework

Answers

C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

What is center of mass?

The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.

What is center of gravity?

Center of gravity is the point from which the weight of a body or system may be considered to act.

Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

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Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .

Answers

The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

Speed of the satellite

v = √GM/r

where;

M is mass of EarthG is universal gravitation constantr is distance from center of Earth = Radius of earth + 4930 km

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

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A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.600 for red light of wavelength 690 nm. If a beam containing these two colors is incident at an angle of 30.0° on a piece of this glass, what is the angle between the two beams inside the glass?

Answers

The angle between the two beams inside the glass is mathematically given as

<(BR)= 0.563°

What is the angle between the two beams inside the glass?

Generally, the equation for an angle for blue is  mathematically given as

[tex]< B= arcsin( sin\theta / i )[/tex]

Therefore

<B= arcsin( sin30 / 1.660 )

<B= 17.53°

For angle for red

<R= arcsin(sin30 / 1.610)

<R= 18.09°

For the angle in between

<(BR)= 18.093 - 17.530

<(BR)= 0.563°

In conclusion,  the angle between the two beams inside the glass

<(BR)= 0.563°

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The motor of a weed trimmer spins
at 9,000 rpm. The amount of time
required for the motor to reach this
speed would NOT be affected by...
A. the distribution of
trimmer line inside the
spool
B. the mass of the spool
containing the trimmer
line
C. the direction in which
the torque is applied

Answers

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A

What is a weed trimmer?

The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.

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A boat moves through the water with two forces acting on it. One is a 1,575-N forward push by the water on the propeller, and the other is a 1,200-N resistive force due to the water around the bow. (Review attachment)

(a) What is the acceleration of the 1,100-kg boat?

______ m/s2

(b) If it starts from rest, how far will the boat move in 20.0 s?

______ m


(c) If it starts from rest, how far will the boat move in 20.0 s?

______ m/s

Answers

(a) The acceleration of the 1,100-kg boat is 0.341 m/s².

(b) The distance covered by the boat is 68.2 m.

(c) The speed of the boat is 6.82 m/s.

Acceleration of the boat

Net force on the boat = 1,575 N - 1,200 N = 375 N

F(net) = ma

a = F(net)/m

a = 375/1100

a = 0.341 m/s²

Distance moved in 20 s

s = ut + ¹/₂at²

s = 0 + ¹/₂(0.341)(20)²

s = 68.2 m

Speed of the boat in 20 s

v = u + at

v = 0 + 0.341(20)

v = 6.82 m/s

Thus, the acceleration of the 1,100-kg boat is 0.341 m/s², the distance covered by the boat is 68.2 m and the speed of the boat is 6.82 m/s.

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describe one similarity and one difference between the velocity on the reference circle and the velocity on the pendulum

Answers

SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.

A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.

What is the velocity at the bottom of a pendulum?

As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)

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Which of the following is an example of an offensive action?

Answers

An example of an offensive action is B. kicking a ball into the goal

Something that is offensive upsets or embarrasses people because it is rude or insulting.

What is a example of an offensive action?

Offensive is an arranged campaign or plan of action, normally created by the military or planned to achieve some specific political aim or goal. A military plan to strike is an example of a military offensive. A plan to fight the war on medications is an instance of an offensive.

What are the types of offensive operations?

The four types of offensive operations are move to contact, attack, exploitation, and pursuit.

Leaders direct these offensive operations sequentially and in combination to generate highest combat power and destroy the enemy.

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Checking the what will reduce the possibility of having to rebuild or replaced the engine?

Answers

Checking the oil will reduce the possibility of having to rebuild or replaced the engine.

what is an engine ?

A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.

The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.

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Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as​

Answers

according to definition of acceleration

a=v-u/t

t=v-u/a(equation 1)

according to the formula of average velocity

v+u/2*s/t

s=v+u/2*t(equation 2)

now putting the value of t in equation 2

s=v+u/2*v-u/a

s=v^2-u^2/2a

v^2=u^2+2as

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Please show work if possible! Thank you!!
A 2.0 x 103 kg roller coaster travels around a vertical 24-m radius loop. If the coaster has a tangential speed of 18 m/s at the lowest point of the loop, what is the normal force that is exerted on the coaster by the track at this point?
a. 5.3 x 10^4 N
b. 4.7 x 10^4 N
c. 3.0 x 10^4 N
d. 2.7 x 10^4 N

Answers

B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

Normal force exerted on the coaster at the lowest point

Fₙ = mg + mv²/r

where;

m is mass of the coasterv is speed of the coasterr is radius of the path

Fₙ = (2,000 x 9.8) + (2,000 x 18²)/24

Fₙ = 46,600 N

Fₙ =  4.7 x 10⁴ N

Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal..
1. What is the horizontal component of the force exerted by the hi.nge on the beam? (Use the `to the right' as + for the horizontal direction.)
2. What is the magnitude of the force that the beam exerts on the hi.nge?

Answers

The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Tension in the cable

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (40 x 9.8)/(2 x sin31)

T = 380.55 N

Vertical component of the force

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

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It's a snowy day and you're pulling a friend along a
level road on a sled. You've both been taking physics,
so she asks what you think the coefficient of friction
between the sled and the snow is. You've been
walking at a steady 1.5 m/s, and the rope pulls up
on the sled at a 35 ° angle. You estimate that the
mass of the sled, with your friend on it, is 57 kg and
that you're pulling with a force of 75 N

Answers

The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

How to find the coefficient of friction between the sled and the snow?Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.To solve the problem, we have to draw the free body diagram of the given system.We have given with the following values,

                                     [tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]

Here, acceleration will be equal to zero, because the velocity is given as constant.

Thus, from the diagram, we can write the balancing equations as follows,

                                      [tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]

Substituting N in f and f in the equation of ma, then we get,

                   [tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]

Substituting values, we get the coefficient of friction as,

                    [tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

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The sled's coefficient of friction with the snow is 0.119.

We must understand the friction in order to choose the solution.

How can I determine the sled and snow's coefficient of friction?A force that works parallel to the surface of contact and opposes the relative motion is present whenever one body moves over the surface of another body. Friction is the name for this opposing force.We must create the given system's free body diagram in order to solve the issue.The values that we have provided are

                               [tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]

Because the velocity is specified as constant in this case, the acceleration will be equal to zero.

Consequently, we can express the balancing equations as follows using the diagram:

                             [tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]

When we substitute N for f and f in the equation for ma, we obtain,

                       [tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]

By substituting values, we obtain the friction coefficient as.

                                   [tex]k=0.119[/tex]

As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.

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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)

Answers

The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

What is acceleration?

Acceleration is the change in velocity with time.

The maximum acceleration is obtained by taking moments about the tipping point of rotation.

Moment = force * perpendicular distanceClockwise moment = Anticlockwise moment

F₂ * 1.58 m = F₁ * 0.67 m

where

F₂ is tipping force = mass * acceleration, a

F₁ is weight = mass * acceleration due to gravity, g

The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m

Height of refrigerator is 158 cm 0r 1.58 cm

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.

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Based on the way living things are organized ehat level combines to form organ ststems

Answers

Answer:

Higher levels of organization are built from lower levels.

Molecules combine to form cells.

Cells combine to form tissues.

Tissues combine to form organs.

Organs combine to form organ systems,

and organ systems combine to form organisms.

Explanation:

Hope it helps.

What is the fastest possible speed called in our universe and what is the equation for it?

Answers

Answer:

The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.

[tex]s\frac{d}{t}[/tex]

Explanation:

hope this helps you my friend

The speed of light. Nothing can travel faster than speed because of their massless particles that include photons. The equation is c = 1/(e0m0)1/2 = 2.998 X 108m/s

When this surgical procedure is used to reduce the risk of stroke it will correct stenosis in the artery the most common cause of this condition is the buildup of plaque that forms in the artery name this procedure

Answers

Endarterectomy is used to reduce the risk of stroke and correct the stenosis in the artery.

What is carotid artery stenosis?

The primary blood vessels that supply the brain with blood and oxygen are the carotid arteries.

The narrowing of these arteries is referred to as carotid artery disease. Carotid artery stenosis is another name for it. The main factor causing constriction is atherosclerosis.

This fat deposit reduces the blood flow to the brain which cause a stroke.

Carotid endarterectomy is a surgical treatment to remove plaque, an accumulation of fatty deposits that causes a carotid artery to become narrowed.

Hence, an Endarterectomy is used to reduce the risk of stroke, it will correct stenosis in the artery.

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Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be larger than the magnitude of A - B by the factor n, what must be the angle between them?​

Answers

Answer:

[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]

Explanation:

1) for A+B: a²+b²-2abcos(π-α);

2) for A-B: a²+b²-2abcos(α);

3) according to the condition (A+B):(A-B)=n, then

[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]

Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?

Answers

Answer:  Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure. The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). Then, at an angle 135.48°, the planet Y rotated through during this time.

Explanation: To find the answer, we need to know about the Kepler's third law of planetary motion.

What is Kepler's third law of planetary motion?Kepler's third law of planetary motion states that, the square of the period of revolution is proportional to the cube of the orbital radius of the elliptical path.It can be expressed as,

                                      T² ∝ r³

How to solve the problem?We have given with the ratio of the radii of their orbits as,4:3.planet X rotated through an angle of 88°.thus,

                 [tex]\frac{r_1}{r_2}=\frac{4}{5} \\\frac{T_1}{T_2} =(\frac{r_1}{r_2})^{3/2}\\[/tex]

As we know that,

                  [tex]T=\frac{2\pi }{w}[/tex] where, w is the angular velocity.

Angular displacement is the angle swept by the position vector of a particle in a given interval of time.

                           [tex]\alpha[/tex] =wt.

We can rewrite our equation as,

                   [tex]\frac{T_x}{T_y}=\frac{w_y}{w_x}\\thus,\\\frac{w_y}{w_x}=(\frac{r_1}{r_2})^{3/2}[/tex]

We have to find the angle rotated by planet Y during 5 yrs. So, we can rewrite the above equation in terms of angular displacement.

                     [tex]\frac{\alpha _y}{\alpha _x} = (\frac{r_1}{r_2})^{3/2}\\where,\\\alpha _x=\frac{88^0}{5 yrs} because,\\here, angle \beta_x =88^0.\\[/tex]

Thus, the angle rotated by planet Y during 5 yrs will be  [tex]\beta _y[/tex] =

                     [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} )^{3/2}=\frac{135.48^0}{5yrs} .\\thus,\\\beta _y=135.48^0.[/tex]

Thus, we can conclude that the angle rotated by planet Y during 5 yrs will be 135.48 degrees.

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The planet Y then rotated through at this time at an angle of 135.48°.

In order to understand the solution, we must be familiar with Kepler's third law of planetary motion.

What does the third law of planetary motion by Kepler say?According to Kepler's third law of planetary motion, the elliptical path's orbital radius is proportional to the cube of the square of the revolution's period.It can be stated as follows:

                              T² ∝ r³

How can the issue be resolved?The ratio of their orbital radii that we have provided is 4:3.Planet X rotated at an 88° angle. thus,

                                  [tex]\frac{R_1}{R_2}=\frac{4}{5} \\\frac{T_1}{T_2}=(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

As we are aware,

                                [tex]T=\frac{2\pi }{w}[/tex]

where w is the angle of rotation per time.

The angle that a particle's position vector sweeps over in a specific amount of time is known as the angular displacement.

                                     [tex]\alpha[/tex]=wt.

Our equation can be rewritten as,

                                   [tex]\frac{w_y}{w_x} =(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

We have to find the angle that planet Y rotated at over the course of five years. Consequently, we can express the equation above in terms of angular displacement.

                                   [tex]\frac{\alpha _y}{\alpha _x}=(\frac{4}{3} ) ^{\frac{3}{2} } , where\\\alpha _x=\frac{88}{5yrs} \\[/tex]

So, during a period of five years, planet Y will rotate at an angle,

                           [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} ) ^{\frac{3}{2} }=\frac{135.48}{5yrs}[/tex]

Thus, we may infer that planet Y will revolve at an angle of 135.48 degrees during the course of five years.

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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.

Answers

The tension, T in the cable is equal to 323.5 N.

What is the tension?

Tension is force exerted by a cable or string on another object usually a weight suspended from the cable or string

The tension in the cable is found this:

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

T = 241.68/0.747

T = 323.5 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

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The light beam shown in the figure below makes an angle of = 24.8° with the normal line NN' in the linseed oil. Determine the angles and '. (The refractive index for linseed oil is 1.48.)

Answers

The angle of refraction of the light beam is determined as  16.46 ⁰.

Angle of refraction of the light beam

n = sin i / sin r

where;'

n is refractive indexi is angle of incidencer is angle of refraction

Angle between the ray and the normal = incident angle = 24.8⁰

1.48 = sin (24.8) / (sin r)

sin r = sin (24.8) / (1.48)

sin r = 0.283

r = sin ⁻¹(0.283)

r = 16.46 ⁰

Thus, the angle of refraction of the light beam is determined as  16.46 ⁰.

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Which type of wave interaction is shown in the photo?

Answers

The wave interaction that is shown in the photo is refraction as light moves from air to water.

What is refraction?

Refraction refers to the change in the frequency of a wave and the direction of the wave as it moves from one medium to another. We know that waves makes a body under water to look slightly different than when it is in air.

Thus, the wave interaction that is shown in the photo is refraction as light moves from air to water.

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a quantity of gas is contained in a sealed container of fixed volume the temperature of the gas is increased

Answers

The average kinetic energy of the gas's molecules directly relates to its temperature. Faster moving particles will more frequently and violently hit the container walls. Because of this, the force acting on the container's walls increases, increasing the pressure.

What happens when the temperature of a gas is increased?

We know that temperature is proportional to the average kinetic energy of a sample of gas. The proportionality constant is (2/3)R and R is the gas constant with a value of 0.08206 L atm K-1 mol-1 or 8.3145 J K-1 mol-1.The average kinetic energy and the velocity of the gas particles striking the container walls rise as the temperature rises. As the temperature rises, the pressure must as well since pressure is the force the particles per unit of area exert on the container.The molecules in a gas gain more energy and can move more quickly as it is heated. The pressure will rise, and there will be greater impacts on the container walls. On the other hand, cooling the molecules will cause them to slow down and lower the pressure.

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When the volume of gas increased the KE of gases particles increases and hit the walls of container which increases the pressure on the walls .

What if gas will increase?

The gas expands in a closed container . The molecules strike and hit each other therefore the pressure at the walls increase.

According to ideal gas law , when volume is cnstant the Pressure and temperature both will increase.

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Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars' radius is 3400 km , determine the mass of Mars.

Answers

The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg

Data obtained from the questionAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of Mar (g) = 0.38 × 9.8 = 3.724 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Radius (r) = 3400 Km = 3400 × 1000 = 3400000 mMass (M) =?

How to determine the mass of Mar

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by G

M = gr² / G

M = (3.724 × 3400000²) / 6.67×10¯¹¹

M = 6.45×10²³ Kg

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A 1500 kg aircraft going 35 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 15 m/s after the collision. If they skid for 112.1 m before stopping, how long (in seconds) did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?

Answers

The two aircrafts are skidded for 14.9 s.

To find the answer, we need to know about the Newton's equation of motion.

What is the Newton's equation that relates velocity, distance, acceleration and time?

As per Newton's equation of motion

V²-U²= 2aSV= U+atV= final velocity, U = initial velocity, S = distance, a= acceleration, t= time

What's the acceleration of the aircrafts that skidded from 35 m/s after the collision for 112.1 m before achieving 15 m/s?Here, U = 35 m/s, V = 15m/s, S= 112.1 mSo, 15²- 35²= 2a×112.1= 224.2a

=> a= -1000/224.2= -4.5 m/s²

What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?Here, U = 35 m/s, V=15 m/s a= -4.5 m/s²35= 15+(-4.5)t

=> t= 20/4.5 = 4.4 s

Thus, we can conclude that the two aircrafts are skidded for 4.4 s.

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When using a stream table in a classroom setting what are three factors that can be controlled?

Answers

Answer:

These include the slope of the land, the nature of the land surface, the placement of dams, and the direction of topsoil disturbance as created by farming activities. Materials: Students should work in groups of 3 or 4, or as materials allow.

Explanation:

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Which best describes the results of Becquerel’s experiments?

Answers

Both forming images when placed in their respective places best describes  Becquerel’s experiments.

What is Becquerel’s experiment?

This was conducted by  Henri Becquerel in which he sought to know how uranium salts are affected by light.

He discovered that the salts emits a penetrating radiation and formed an image in the presence of light but didn't form any in darkness.

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