Answer: The statement [tex]T_{1} > T_{2}[/tex] must be true.
Explanation:
As it is given that heat is being transferred from a substance with temperature [tex]T_{1}[/tex] to a substance with temperature [tex]T_{2}[/tex].
It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.
Hence, in the given situation [tex]T_{1} > T_{2}[/tex].
Thus, we can conclude that the statement [tex]T_{1} > T_{2}[/tex] must be true.
An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surface, in terms of g, the acceleration of gravity at Earth's surface. A planet's gravitational acceleration is given by gp = G Mp/r^2p
a. 12.0 g.
b. 48.0 g.
c. 6.00 g.
d. 96.0 g.
e. 24.0 g.
Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
A box is being pulled to the right over a rough surface. t > fk , so the box is speeding up. Suddenly the rope breaks. What happens? The box:_________.
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
b. stops immediately.
c. continues speeding up for a short while, then slows and stops.
d. continues with the speed it had when the rope broke.
Answer:
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
Explanation:
Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.
Since there is a net force, there is an acceleration and thus an increasing velocity.
When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.
Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.
So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.
So, the answer is a.
How is fitness walking beneficial?
It can relieve stress and improve mood.
It can decrease energy levels.
It can decrease perspiration.
It can relieve allergy symptoms.
Answer:
It can relieve stress and improve mood.
A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference between the point at the center of the ring and a point on its axis at a distance 6R from the center. (Use any variable or symbol stated above along with the following as necessary: ke.)
ΔV = V(0) − V(6R) = ?
Answer:
the electric potential difference between the point at the center of the ring and a point on its axis ΔV is [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex]
Explanation:
Given the data in the question;
electric potential at the center of the ring V₀ = kQ / R
electric potential on the axis point Vr = kQ / √( R² + x² )
at a distance 6R from the center,
point at x = 6R
so distance circumference r = √( R² + (6R)² )
so
electric potential on the axis point Vr = kQ / √( R² + (6R)² )
Vr = kQ / R√37
Now
ΔV = V₀ - Vr
we substitute
ΔV = ( kQ / R) - ( kQ / R√37 )
ΔV = kQ/R( 1 - 1/√37 )
ΔV = kQ/R( 1 - 0.164398987 )
ΔV = kQ/R( 0.8356 )
ΔV = [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex] { where k = [tex]\frac{1}{4\pi e_0}[/tex] }
Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex]
A circular wire loop is placed near a long, straight, current-carrying wire in which the current is either increasing or decreasing. Draw the situations in which the induced current in the loop counterclockwise?
Answer:
The induced current is counter clockwise if the current is decreasing and towards right.
Explanation:
When the current is decreasing in the wire, the direction of magnetic field at the center of the loop is outwards to the plane of paper which is given by the Maxwell's right hand thumb rule. The magnetic field is decreasing in nature.
So according to the Lenz's law, the induced current is such that which opposes the cause of its production, so that the induced current is counter clockwise.
a ball is projected horizontally from the top of a hill with a velocity of 30 m per second if it reaches the ground 5 seconds later the height of the hill is
Answer:
The height of the hill is 125m.
Explanation:
Since the ball is projected horizontally from the top of a hill, there is no vertical component of the velocity of projection. Therefore, so far as motion in the vertical direction is concerned, the ball is just dropped, ie its initial velocity u is merely 0 m/s. It reaches the ground after 5 seconds from the moment of projection, under the action of accerelation due to gravity.
Using,
s = u t + ½ a t². In this expression u= 0 m/s, a = 10m/s², t = 5 s. Substituting in the equation we get,
s = 0× t + ½ ×10 m/s²× 5²s²= 5× 25 m = 125m.
Instantaneous speed is...
a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.
PLEASE HURRY
Answer:
The speed attained at a particular instant in time.
Explanation:
Instantaneous speed is the speed attained at a particular instant in time.
It is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
It is equal to the rate of change of speed.
It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.
Hence, the correct option is (b).
Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))
Explanation:
it depens on the subject and object. Let's example
you are driving a tesla car with your dog sitting on your side. You will say that the dog is at REST
but your friend, standing in sidewalk, seeing the same dog, will say that your dog is moving because it has MOTION from your car
What is the magnitude of the resultant velocity for a bird flying first at a
speed of 10 m/s North East and then flying to South at a speed of 8 m/s?
Answer:
the magnitude of resultant is landify space ociured by meteroid
The magnitude of resultant velocity for a bird flying is 7.132 m/s.
What is Velocity?Velocity is defined as the directional motion of an object which is observed from a particular frame of reference and measured by a particular standard of time, as an indication of the rate of change of position.
Velocity is the Vector quantity as it has magnitude as well as direction. It can be represented as
[tex]\overline{v}={\frac{\Delta x}{\Delta t}}[/tex]
[tex]\overline{v} = average velocity[/tex]
[tex]{\Delta x} = displacement[/tex]
[tex]{\Delta t} = change in time[/tex]
For the information above, two vectors are represented as the sides of a parallelogram, then the resulting vector is given as the diagonal of that parallelogram. The resultant vector is that it can have any value between a maximum and a minimum depending on the angle; But scalar numbers can have either maximum or minimum value depending on whether they are added or subtracted.
The resultant vector of any two vectors is given by: [tex]v^2= \sqrt{v_1^2+ v_2^2+ 2 v_1v_2 cos }[/tex]Θ
where, Θ is the angle between these two vectors.
The angle between directions north-east and south is 135°.
So, the magnitude of the resultant vector is
[tex]v= \sqrt{10^2 + 8^2 +2.10.8. cos 135}[/tex]
v= 7.132 m/s
Thus, the magnitude of resultant velocity for a bird flying is 7.132 m/s.
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A clay vase on a potter's wheel experiences an angular acceleration of 7.24 rad/s2 due to the application of a 13.3-N m net torque. Find the total moment of inertia of the vase and potter's wheel.
Answer:
[tex]I=1.83\ kg-m^2[/tex]
Explanation:
Given that,
The angular acceleration of the wheel,[tex]\alpha =7.24\ rad/s^2[/tex]
Net torque,[tex]\tau=13.3\ N-m[/tex]
We need to find the total moment of inertia of the vase and potter's wheel. We know that,
Net torque,[tex]\tau=I\alpha[/tex]
Where
I is the moment of inertia
So,
[tex]I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{13.3}{7.24}\\\\I=1.83\ kg-m^2[/tex]
So, the moment of inertia of the vase is equal to [tex]1.83\ kg-m^2[/tex].
Which of the following is a good example of a contact force?
ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up
Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
A reaction requires 1.5 mL of ammonia if it occurs at 1.65 atm and 23 degrees celsius. If the temperature is changed to 30 degrees celsius, what will the new pressure be if the volume remains the same?
Answer: The new pressure will be 1.69 atm
Explanation:
Gay-Lussac's law states that the pressure of the gas is directly proportional to the temperature of the gas at constant volume and the number of moles.
Mathematically,
[tex]P\propto T[/tex] (At constant volume and number of moles)
OR
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] .....(1)
[tex]P_1\text{ and }T_1[/tex] are the pressure and temperature of the gas
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas
We are given:
[tex]P_1=1.65atm\\T_1=23^oC=[23+273]K=296K\\V_2=?mL\\T_2=30^oC=[30+273]K=303K[/tex]
Putting values in equation 1, we get:
[tex]\frac{1.65atm}{296K}=\frac{P_2}{303K}\\\\P_2=\frac{1.65\times 303}{296}\\\\P_2=1.69atm[/tex]
Hence, the new pressure will be 1.69 atm
You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet.
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
Answer:
1. 5.825 × 10¹⁷ J
2. i. $ 29.125 billion ii. I would not invest in the company
3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. a i. The momentum will not be relativistic
ii. This is because objects with large masses do not move at relativistic speeds
b i. 155 m/s
ii. This speed wouldn't be a problem for the ship.
Explanation:
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)
So, K = (γ - 1)mc²
= (1/√(1 - β²) - 1)mc²
= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²
= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²
= 11.65 × 10¹⁶ kgm²/s²
= 1.165 × 10¹⁷ J
Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have
20 % of E = K
0.2E = K
E = K/0.2
= 1.165 × 10¹⁷ J/0.2
= 5.825 × 10¹⁷ J
2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
i. how much would this launch cost?
Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion
ii. Would you invest in this company?
I would not invest in the company
3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy, then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
i. Will the recoil momentum of the ship be relativistic?
The momentum will not be relativistic.
ii. Justify your argument.
This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
i. At what speed will the ship recoil after it launches a probe?
Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J
Also, E'² = (pc)² + (mc²)² where p = momentum of payload
So, making p subject of the formula, we have
(pc)² = E'² - (mc²)²
pc = √[E'² - (mc²)²]
p = √[E'² - (mc²)²]/c
substituting the values of the variables into the equation, we have
p = √[E'² - (mc²)²]/c
p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s
p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s
p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s
p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s
p = 18.59 × 10¹⁶/3 × 10⁸ m/s
p = 6.20 × 10⁸ kgm/s
From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.
So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship
So, v = p/m'
= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg
= 1.55 × 10² m/s
= 155 m/s
ii. Do you think that this is a problem for the ship?
Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.
Explain the meaning of work done and give examples
Answer:
the situation in which the force or any kinds of things rhat are done by force is called work done.ex.typed,wrote
Answer:
WORK DONE on a body is the product ofFORCE IN THE DIRECTION OF DISPLACEMENT AMD DISPLACEMENT.WORK ONLY HAS MAGNITUDE AND NO DIRECTION.
FORCE applied IN the direction of displacement × displacement = POSITIVE WORK DONEFORCE applied OPPOSITE to the direction of displacement × displacement = NEGATIVE WORK DONEFORCE applied PERPENDICULAR to the direction of displacement × displacement = ZERO WORK DONEhope it helps
have a nice day
3) show thal escape veloerty ve where symbols have their usual meaning?
Answer:
[tex]v = \sqrt{\frac{2GM}{r} }[/tex]
The formula for escape velocity where:
G - Gravitational constant (9.81 etc.)
M - the mass of the object the escape should be made from
r - distance to the centre of that mass
An Na ion has a charge of +1 and an Oion has a charge of -2. How many Na ions does one Oion need to balance
charges?
If we warm a volume of air, it expands. Does it then follow that if we expand a volume of air, it warms? Explain.
Answer:
nope don't think so
Explanation:
the heat causes the molecules to move faster therefore expanding in watever it the air is in
Answer:
If the pressure is maintained at a constant value then both statements are equivalent:
P V = n R T Ideal gas equation
if P, n, and R are maintained at constant values then
V = T and the two expressions are equivalent
What relationship do you see between a star colour and temperature
Answer:
Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is
Answer:
Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature – the hotter the star, the more blue it is; the cooler the star, the more red it is.
Explanation:
Pa brainliest po ty
Describe an experiment to find the density of copper turning using a density bottle and kerosene
The density is the ratio of mass to volume of a substance.
What is the density bottle?The density bottle is used to obtain the density of substance by measuring the volume of the fluid displaced.
If the mass of copper turnings are previously weighed and known, the volume of the fluid displaced in the density bottle is the volume of the copper turning.
Hence;
Density = mass/ volume
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Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Explanatio
omega=2pi/T
Answer:
0
0000
Explanation:
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter cup containing 844 g of water at 14.6°C?
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
[tex]m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a[/tex]
where,
[tex]m_c[/tex] = mass of copper = 227 g
[tex]m_w[/tex] = mass of water = 844 g
[tex]m_a[/tex] = mass of aluminum = 155 g
[tex]C_c[/tex] = specific heat capacity of calorimeter = 385 J/kg.°C
[tex]C_w[/tex] = specific heat capacity of water = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum = 890 J/kg.°C
[tex]\Delta T_c[/tex] = change in temperature of copper = 283°C - T
[tex]\Delta T_w[/tex] = change in temperature of water = T - 14.6°C
[tex]\Delta T_a[/tex] = change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
[tex](227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}[/tex]
T = 20.84°C
40 ohms
1.2 A
40 ohms
12 V
Calculate the total energy developed in 5
minutes by the system above.
Answer:
17280 J and 1080 J
Explanation:
Given :
R= 40 ohm
I=1.2A
t= 5 min=60×5=300 sec
Now,
Total energy can be calculated as:
[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]
Now,
V=12V
R=40 Ohm
[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]
Total energy is 17280 J and 1080 J
A 1,500-kg truck has a net force of 4,200 N acting on it . What is the trucks' acceleration
Answer:
2.8 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
f is the force
m is the mass
From the question we have
[tex]a = \frac{4200}{1500} = \frac{42}{15} \\ = 2.8[/tex]
We have the final answer as
2.8 m/s²Hope this helps you
which describes a homogeneous mixture
Answer:
A homogeneous mixture is a type of mixture in which the composition is uniform and every part of the solution has the same properties. Example, air
Explanation:
Two oppositely charged objects are separated by a small distance. The objects are then moved three times farther apart from each other. Which of the following statements best describes what happens to the electrical force between the objects?
A: The force of attraction increases
B: The force of attraction decreases
C: The force of attraction stays the same
D: The force of attraction becomes a force of repulsion
The statement best describes what happens to the electrical force between the objects when moved three times farther apart from each other is the force of attraction decreases. The correct option is B.
What is force of attraction?
The objects when placed at a distance, they are attracted by the gravitational force of attraction.
Gravitational force of attraction is directly proportional to the product of masses and inversely proportional to the square of distance between them.
Two oppositely charged objects are separated by a small distance. The objects are then moved three times farther apart from each other.
Force is indirectly proportional to the distance. So, when the distance increases the force of attraction decreases.
Thus, the correct option is B.
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The phrase "hot air rises" describes which type of heat transfer?
A. Conduction
B. Insulation
C. Convection
D. Radiation
Answer:
Option C
Explanation:
C. Convection...
The phrase 'hot air rises'describe the convection as a type of heat transfer therefore the correct answer is option C.
It is the type of heat transfer in which the heat is transferred from the bulk motion of the particles of liquid and gases due to the presence of temperature gradient.
The hot air rises is one of the best example of convection currents
The boiling water is also an example of convection heat transfer process in which the hot water rises to the top of the surface due to the bulk molecular movement of convection current .
The hot air rides is the bulk movement of the gas molecules because of the presence of the temperature gradients resulting in the varied density of air ,the hot air rising to the surface have less density as compared to the air settling near the surface ,the convection current phenomenon is responsible for the phrase "hot air rises" ,therefore the correct option is the option C.
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A 10,000J battery is depleted in 2h. What power consumption is this? *
A) 5000W
B) 3W
C) 1.4W
D) 20000W
show your work please
Answer:
C) 1.4W
Explanation:
Given;
energy of the battery, E = 10,000 J
duration of the battery, t = 2 hours
The power consumption of the battery is calculated as;
[tex]Power = \frac{Energy}{time} \\\\Power = \frac{10,000}{2 \times 3600 } \\\\Power = 1.4 \ J/s = 1.4 \ W[/tex]
Therefore, the power consumption of the battery is 1.4 W
a camera employs _lens to form_images
Answer:
a camera employs camera lens to firm some images.
Explanation:
hope this helps.
a car runs of a road and collides with a tree. glass pieces from the windscreen are projected forward and are found an average distance of 12m from the car. the average height of the windscreen is 1.2m.
Establish the speed of the car at the time of impact. assume g=10 m/s²
Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
Hooke's law describes a certain light spring of unstretched length 38.0 cm. When one end is attached to the top of a door frame and a 6.00-kg object is hung from the other end, the length of the spring is 42.0 cm.
a. Find its spring constant.
b. The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.
Answer:
(a) 1470 N/m
(b) 48.2 m
Explanation:
Applying,
(a) F = ke.................... Equation 1
Where F = force applied to the spring, k = spring constant, e = extension
make k the subject of the equation
k = F/e............... Equation 2
But,
F = mg............. Equation 3
Where m = mass, g = acceleration due to gravity
Substitute equation 3 into equation 2
k = mg/e.............. Equation 4
From the question,
Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m
Constant: g = 9.8 m/s²
Substitute these values into equation 4
k = (6×9.8)/0.04
k = 1470 N/m
(b) Consider the end of the spring to the left which exert a force to the right
Then,
e = F/k............. Equation 5
Given: F = 150 N, k = 1470 N/m
Substitute these values into equation 5
e = 150/1470
e = 0.102 m
Hence the length of the spring is
L = 0.38+0.102 = 0.482 cm = 48.2 m