A grasshopper jumps at a 63.0 angle with an initial velocity of 4.22 m/s. how far away does it land

Answers

Answer 1

The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

To calculate the distance of the landing point of the grasshopper to the point where its jumps, we use the formula of range.

What is horizontal range?

Range can be defined as the horizontal distance between the point of projection to the point where the projectile hit the plain again.

R = u²sin2∅/g........... Equation 1

Where:

R = Distance between the point of jump and the point at which it landsu = initial velocity∅ = angleg = acceleration due to gravity

From the question,

Given:

u = 4.22 m/s∅ = 63°g = 9.8 m/s²

Substitute these values into equation 1

R = (4.22)²sin(2×63)/9.8R = 1.47 m

Hence, The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

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Related Questions

A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per meter of length. By constructing a suitable Gaussian surface around the wire, Find the magnitude and direction of the electric field at points

a)5.5m

b)2.5m

perpendicular from the center of the wire.​

Answers

(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

Electric field at a point on the Gaussian surface

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

λ is linear charge densityε₀ is permitivity of free spacer is the position of the chargeAt a distance of 5.5 m

[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C[/tex]

At a distance of 2.5 m

[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C[/tex]

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

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A 4500 kg Aston Martin traveling at 102 m/s has to stop short because some ducklings
hazard onto the road. The Aston Martin was able to stop in 1.77 seconds. How much
force was placed on the car?

Answers

Answer:

-259322.03N

Explanation:

[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]

Two identical charges are located 1 m apart and feel a 1 N repulsive electric force. What is the charge of each particle.

Answers

The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C

Assumption

Let the charge on each particles be q

How to determine the charge Final force (F) = 1 NDistance apart (r) = 1 mElectrical constant (K) = 9×10⁹ Nm²/C²Charge on 1st particle (q₁) = q =? Charge on 2nd particle (q₂) = q =?

The charge on each particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = Kq² / r²

1 = (9×10⁹ × q²) / 1²

1 = 9×10⁹ × q²

Divide both side by 9×10⁹

q² = 1 / 9×10⁹

Take the square root of both side

q =  √(1 / 9×10⁹)

q = 1.05×10¯⁵ C

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A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA hangs vertically from the pulley. The mass is released from rest. show answer Incorrect Answer 25% Part (a) What is the magnitude of the tangential acceleration of the hanging mass?

Answers

The magnitude of the tangential acceleration of the hanging mass is 2mg/MR

Tangential acceleration of the hanging mass

The tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;

τ = Iα

Where;

I is the moment of inertiaα is the angular velocity

[tex]\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}[/tex]

Where;

m is the hanging massM is the mass of solid disk

The tangential acceleration is calculated as follows;

[tex]a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}[/tex]

Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR

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. Radiation travels at the speed of light
T or F?

Answers

Answer:

electromagnetic radiation moves at the speed of light

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.

Answers

Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.

Please answer and show formula

Answers

Voltage = 5 volts

[tex]\sf \dfrac{number \ of \ turns \ in \ primary \ coil}{voltage\ in \ primary \ coil} = \dfrac{number \ of \ turns \ in \ secondary\ coil}{voltage\ in \ secondary\ coil}[/tex]

[tex]\hookrightarrow \sf \dfrac{400}{100} = \dfrac{20}{V_2}[/tex]

[tex]\hookrightarrow \sf 400V_2}{} = 20*100[/tex]

[tex]\hookrightarrow \sf V_2 = 5 V[/tex]

what is one type of compact star with a mass similar to the sun but a diameter similar to earth?

Answers

Explanation:

neutron star, any of a class of extremely dense, compact stars thought to be composed primarily of neutrons. Neutron stars are typically about 20 km (12 miles) in diameter. Their masses range between 1.18 and 1.97 times that of the Sun, but most are 1.35 times that of the Sun.

A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after being lifted 2.00 m ? Solve this problem using work and energy

Answers

The final speed of the block, after being lifted 2.00 m is 3.39 m/s

What is speed?

Speed can be defined as the rate of change in the distance of a body.

To calculate the speed of the block after being lifted 2.00 m,  first, we need to calculate the acceleration of the block using the formula below

Formula:

T-mg = ma......... Equation 1

Where:

T = Tension in the cablem = mass of the cablea = accelerationg = acceleration due to gravity

Restructuring the formula above,

a = (T-mg)/m............... Equation 2

From the question,

Given:

T = 19 Nm = 1.5 kgg = 9.8 m/s²

Substitute these values into equation 2

a = [(19)-(1.5×9.8)]/1.5a = 4.3/1.5a = 2.87 m/s²

Finally, to calculate the speed of the block, we use the formula below.

v² = u²+2as.......... Equation 3

Where:

v = Final speedu = initial speeda = accelerations = distance

From the question,

Given:

u = 0 m/sa = 2.87 m/s²s = 2.00 m

Substitute these values into equation 3

v² = 0²+(2×2×2.87)v² = 11.48v = √11.48v = 3.39 m/s

Hence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.

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Water of mass 3 kg at a temperature of 80 ℃ is added to 5 Kg of water at 5 ℃. Calculate the final temperature of the mixture

Answers

The final temperature of the mixture is 33.123 °C.

What is temperature?

Temperature can be defined as the hotness or coldness of a thing or place.

To calculate the final temperature of the mixture, we use the formula below.

Formula:

Heat gained by the cold water = heat lost by the hot watercm(t₃-t₁) = cm'(t₂-t₃)m((t₃-t₁) = m'(t₂-t₃)......... Equation 1

Where:

m = mass of the cold waterm' = mass of the hot watert₁ = Temperature of the cold watert₂ = Temperature of the hot watert₃ = Temperature of the mixture.

make t₃ the subject of the equation

t₃ = (mt₁+m't₂)/(m+m')............. Equation 2

From the question,

Given:

m = 5 kgm' = 3kgt₁ = 5 °Ct₂ = 80 °C

Substitute these values into equation 2

t₃ = [(5×5)+(3×80)]/(3+5)t₃ = (25+240)/8t₃ = 33.123 °C

Hence, The final temperature of the mixture is 33.123 °C.

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Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?

Answers

There is no temperature change which drives heat flow, thus no heat will be released by the water.

Heat released by the water when it freezes

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - Фi

Initial temperature of water, Фi = 0 °C

when water freezes, the final temperature, Фf = 0 °C

Q = 22 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

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Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.
Express your answer in terms of given quantities and ϵ0.
View Available Hint(s)for Part A


F(V)

Answers

The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

Magnitude of the force

The magnitude of the force each plate experiences due to the other plate is determined as follows;

F = U/d

where;

U is potential energy stored in the capacitor

[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]

Q = CV

[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]

where;

C is the capacitance

The capacitance is given as;

[tex]C = \frac{\varepsilon _o A }{d}[/tex]

[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]

Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

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list out the use of simple machine​

Answers

Explanation:

simple machine can multiplayer of speed and force

A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed 4.20 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 8.40 block recoils with a speed of 0.400 m/s. In the figure; the blocks are in contact for 0.200 s.

Answers

For A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed of 4.20 m/s  is mathematically given as

F = 193.2N

What is the magnitude of the average force on the 8.40-kg block, while the two blocks are in contact, is closest to?

Generally, the equation for the  magnitude of the average force mathematically given as

F = m(v1+v2)/t

F = 8.40(4.2+O.4)/t

F = 193.2N

In conclusion magnitude of the average force is

F = 193.2N

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If a transverse wave osculates 7 times every second and the speed of the wave is 27 m/s what is the wavelength of the wave

Answers

Explanation:

formula is ˠ=vf

f=1/T

1/7

f=0.14

wavelength=27Ⅹ0.14

=3.78m

OR

7Ⅹ27

=189m

Mention the objective of the Experiment?

Answers

Answer:

I don't understand your question ❓,the object.....of what experiment

The objective of this experiment is to learn whether any Brainly contributor can give a reasonable or helpful answer to a question that contains totally zero information.

Q. 1 MWH is equal to ------- joules
a.3.6*10^10
b.3.6*10^6
c.3.6*10^9
d.3.6

Answers

Correct option is B
1 kWh is equal to 3.6 × 10
6

1 Watt-sec is a Joule a watt expended for a second.
Because there’s 3600 seconds in an hour,
then 3600 Ws = 1 Wh = 3600 Joules = 3.6 kJ
so 1000 Wh = 1 kWh = 1000 × 3.6 kJ = 3.6 MJ = 3.6 × 10
6

17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus

Answers

Answer: b or d

Explanation: b or d

Answer:

a

Explanation:

F= ma

interestingly

when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases

(man, PHYSICS IS JUST THE BESY)

A: a

object with the smallest mass has largest acceleration

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,
a. Sketch the signal for -4 b. Find the fundamental period of the signal (if it is periodic).
c. Determine if the signal is odd / even or neither.
d. Compute the energy of the signal for all time.
e. Compute the power of the signal for all time.

Answers

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

How can the factors of the wave be found?

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = 4

c. The signal is even, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

[tex] \frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda[/tex]

Which gives;

E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg

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Can someone please help me with this assignment, this is due today

Answers

Answer:

did you get it done if not lmk I will help you out tomorrow when I get up

_______ is an SI unit for mass.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight

Answers

Answer:

B. kilogram

I hope this helps you

:)

Answer:

B. kilogram

Explanation:

When stated in the unit J s, which is equal to kg m2 s1, the kilogram (kg) is defined by considering the fixed numerical value of the Planck constant h to be 6.62607015 1034 when expressed in the unit J s, which is equivalent to kg m2 s1. The United States Prototype Kilogram 20, a platinum-iridium cylinder held at NIST, is the country's principal mass standard. The kilogram was initially known as the Kilogram of the Archives, and it was defined as the mass of one cubic decimeter of water at its greatest density temperature. It was superseded by the International Prototype Kilogram following the International Metric Convention in 1875, which became the unit of mass without reference to the mass of a cubic decimeter of water or the Archives Kilogram. National Prototype Meters and Kilograms were allocated to each country that signed the International Metric Convention. Learn more about the kilogram's history and current definition. The kilogram (kg) is the only SI basic unit whose name and symbol incorporate a prefix for historical reasons. The SI prefix for 1000 or 103 is "kilo." Prefix names and symbols are attached to the unit name "gram," and prefix symbols are attached to the unit symbol "g," to create names and symbols for decimal multiples and submultiples of the unit of mass. Find out more about this historical oddity.

Units of Mass

10 milligrams (mg) = 1 centigrams (cg)

10 centigrams = 1 decigrams (dg) = 100 milligrams

10 decigrams = 1 gram (g)

10 decigrams = 1000 milligrams

10 grams = 1 dekagrams (dag)

10 dekagrams = 1 hectogram (hg)

10 dekagrams = 100 grams

10 hectograms = 1 kilogram (kg)

10 hectograms = 1000 grams

1000 kilograms = 1 megagram (Mg) or 1 metric ton (t)

A body's mass is a measurement of its inertial property, or the amount of stuff it contains. The force imposed on a body by gravity or the force required to maintain it is measured by its weight. On Earth, gravity accelerates a body downward at around 9.8 m/s2. In the context of weights and measurements, weight is frequently used as a synonym for mass. The verb "to weigh," for example, meaning "to ascertain the mass of" or "to have a mass of." Weight should be phased out in favor of mass, and the term mass should be used when mass is indicated. The kilogram is the SI unit of mass (kg). The weight of a body in a given reference frame is defined in science and technology as the force that causes the body to accelerate at the same rate as the local acceleration of free fall in that reference frame. As a result, the newton is the SI unit for the amount weight defined in this way (force) (N).

a=5i+4j-6k ,b=-2i+2j+3k ,c=4i+3j+2k. find the vector perpendicular to a and c​

Answers

Answer:

Explanation:

You can use the cross product. Let the vector that perpendicular to a and c is [tex]\vec{d}[/tex], so:

[tex]\vec{d}=\vec{a}\times\vec{c}=\left|\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\5&4&-6\\4&3&2\end{array}\right] \right|=(8+18)\hat{i}-\hat{j}(10+24)+\hat{k}(15-16)=26\hat{i}-34\hat{j}-\hat{k}[/tex]

To check that c is perpendicular with a and b, do the dot product between c and a and also c and b and if the result is zero, you're true.

[tex]\vec{d}.\vec{a}=(26*5)-(34*4)+(6)=0[/tex]  (c perpendicular to a)

[tex]\vec{d}.\vec{c}=(4*26)-(34*3)-(2*1)=0[/tex] (d perpendicular to c)

An electron has a mass of 9.1x10^-31 kg and a charge
of -1.6x10^-19 C. Suppose you could isolate one electron in
a perfect vacuum and then create an electric field to pull
and edi 107. upward on the electron. How strong would the field have to
be to counteract the electron's weight? (In other words,
di Delfine bu how strong would the field have to be to put the electron in
a state of force equilibrium?)

Answers

is one of these the question

2. A tennis ball machine launches balls horizontally with an initial speed of 5.3 m/s, from a height of 1.2 m.
a) What will the time of flight be for a tennis ball launched by the ball machine? (3)
b) What will the range of the tennis ball be? (2)
c) What will be the final velocity of the ball with which it reaches the ground? (3)

Answers

(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

Time of flight of tennis ball

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

Range of the tennis ball

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

Final velocity of the ball

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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Draw free body diagrams for the following objects: (12pts)
1A) A coaster sitting under a cup of coffee.
1B) A car slowing down as it approaches a stop sign.
1C) Your test stuck to your fridge by a magnet.
1D) A baseball just before it leaves the bat.

Answers

(a) The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

(b) The force diagram of a car slowing down as it approaches a stop sign includes force of the car and frictional force opposing the motion.

(c) The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

(d) The force diagram of baseball before it leaves the bat incudes only the weight of the baseball acting downwards.

Force diagram of coaster sitting under a cup of coffee

The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

                                        ↑ Fn

                                        Ф                  Fn = W

                                        ↓ W

Where;

W is weight of the coaster plus weight of coffeFn is the normal reactionForce diagram of a car slowing down as it approaches a stop sign

The force diagram inlcudes the applied force and frictional force opposing the motion.

                                      Ff ←  Ф  → F

where;

Ff is the kinetic frictional forceF is force of the car

Force diagram of test stuck to your fridge

The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

                   

                               Fb ←  Ф → Fa

where;

Fa is the force of the testFb is the force of the fridge

Force diagram of baseball before it leaves the bat

The force diagram includes only the weight of the baseball acting downwards.

                                 Ф

                                  ↓

                                 W = mg

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a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning

Answers

Answer:

Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.

Explanation:

A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap

Answers

Answer:

A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap

Find the temperature

Answers

Answer:

-------1

Explanation:

beacuse that is what i know

As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?

Answers

As a conservation biologist for the Chesapeake Bay, you and your

colleagues have been conducting a research study that tracks the decrease

in the bald eagle population over the past few years.

What evidence can you find for the decrease in the bald eagle population?

Determine
i. the total capacitance for the circuit
ii. the total charge stored in the circuit
iii. the charge stored in C9 (3μF)​

Answers

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

Total capacitance of the circuit

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

C1 and C2 are in series

[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]

C1 and C2 are parallel to C3

[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]

C(123) is series to C5 and C6

[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]

C7 and C8 are in series

[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]

Total capaciatnce of the circuit

Ct + C(78) = 2 μF + 3 μF = 5 μF

Total charge stored in the circuit

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

Charge stored in 3μF capacitor

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

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