. Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/s, and the mass of rain per second striking the roof is 0.071 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof.

Answers

Answer 1

Answer:

the average force exerted by the rain on the roof is 1.065 N

Explanation:

Given;

speed of the rainfall, v = 15 m/s

the mass of the rate of the rainfall, m' (m/t) = 0.071 kg

Let the average force exerted by the rain on the roof  = F

The average force exerted by the rain on the roof  is calculated by Newton's second law of motion;

[tex]F = ma = m \frac{v}{t} = \frac{m}{t} \times v \\\\Recall, \ m' = \frac{m}{t} \\\\F = m'\times v\\\\F = 0.071 \times 15\\\\F = 1.065 \ N[/tex]

Therefore, the average force exerted by the rain on the roof is 1.065 N


Related Questions

You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?

Answers

Answer:

44.6 m

Explanation:

From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.

E = E'

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J

Substituting the values of the variables into the equation, we have

U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂

mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J

9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J

(735 kgm/s²)h + 75  kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J

(735 kgm/s²)h + 121.5  kgm²/s² = 29400 kgm²/s² + 3500 J

(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J

(735 kgm/s²)h + 121.5 J = 32900 J

(735 kgm/s²)h = 32900 J - 121.5 J

(735 kgm/s²)h = 32778.5 J

h = 32778.5 J/735 kgm/s²

h = 44.6 m

So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.

g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Answers

Answer:

[tex]V=8.08m/s[/tex]

Explanation:

From the question we are told that:

Height[tex]h=5.00m[/tex]

Mass [tex]m=0.750kg[/tex]

Radius [tex]r=4.00cm=>0.04m[/tex]

Generally the equation for Total energy is mathematically given by

  [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]

Therefore

 [tex]V=\sqrt{\frac{4gh}{3}}[/tex]

 [tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]

 [tex]V=8.08m/s[/tex]

if p=2i+4j+3k and q=I+5j-2k,find P×q.

Answers

Answer:

[tex]p\times q=-23i+7j+6k[/tex]

Explanation:

We are given that

p=2i+4j+3k

q=i+5j-2k

We have to find pxq

We know that

[tex]p\times q=\begin{vmatrix}  i&j  &k\\  2&4  & 3\\  1& 5 & -2\end{vmatrix}[/tex]

[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]

[tex]p\times q=-23i+7j+6k[/tex]

Hence,[tex]p\times q=-23i+7j+6k[/tex]

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

you drove 50km

Explanation:

10×5 hope this helps

Answer:

50 Km

Explanation:

This is how far you have got on your journey if traveling like this.

Please Mark as Brainliest

Hope this Helps

A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)

Answers

Answer:

The force between the 10 th car and the 11 th car is 13636.4 N.

Explanation:

Force, F = 150 kN

acceleration, a = 2 m/s^2

Let the mass of each car is m. \Total numbers of cars = 11

F = n m a

150000 = 11 x m x 2

m = 6818.18 kg

The force between the 10 th and 11 th car is

T = ma = 6818.18 x 2 = 13636.4 N

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time

Answers

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

PLEASE ANSWR 1ST AND I WILL MARK U BRAINLIEST
Two statements are given- one labeled Assertion (A) and the other labeled Reason ®. Select
the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
a. Both A and R are true, and R is correct explanation of the assertion.
b. Both A and R are true, but R is not the correct explanation of the assertion.
c. A is true, but R is false.
d. A is false, but R is true.
Assertion: An object has a negative acceleration.
Reason: The velocity of an object decreases in the same direction.

Answers

Answer:

Where is the R statement?

Two point charges exert a 6.10 N force on each other. What will the force become if the distance between them is increased by a factor of 8

Answers

Answer:

0.0953125 N

Explanation:

Applying,

F = kq'q/r²................. Equation 1

Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.

From the equation,

If both charges remain constant,

Therefore,

F = C/r²

C = Constant =  product of the two charge(q' and q) and k

Fr² = F'r'²................ Equation 2

From the question,

Given: F = 6.10 N

Assume: r = x m, r' = 8x

Substitute these value into equation 2

6.1(x²) = F'(8x)²

F' = 6.1/64

F' = 0.0953125 N

Hence the new force will become 0.0953125 N

What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs

Answers

Organs are made up of different types of organs.

A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.

Answers

This question is incomplete, the complete question is;

Flag

A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

Problem

A charged particle is moving in the presence of uniform magnetic field. The mass of the particle

is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the

beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).

a.) How large will the x component of the velocity of the particle be in t = 2 s?

b.) Where will the particle be in t = 3.14 s?

c.) How large will the magnitude of the velocity be in t = 2.5 s?​

Answers

Answer:

Answer is a I checked the work

If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?

Answers

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W

Answers

Compute the work done on the table:

W = Fd = (320 N) (32 m) = 10,240 J

Divide this by the given time duration to get the power output:

P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W

Find the ratio of the diameter of copper to iron wire, if they have the same resistance per unit length (as they might in household wiring). dCu dFe =

Answers

Answer:

The ratio of diameter of copper wire to the iron wire is 0.42.

Explanation:

length of both the wires is same as L and resistance is R.

resistivity of copper = 1.7 x 10^-8 ohm m

resistivity of iron = 9.7 x 10^-8 ohm m

Let the diameter of copper is d and for iron is d'.

The formula of the resistance is

[tex]R = \rho\times\frac{L}{A}\\\\R = \rho\times\frac{4L}{\pi d^2}.... (1)\\And\\ R = \rho'\times\frac{4L}{\pi d'^2}.... (2)\\\\comparing (1) by (2)\\\\1.7\times10^{-8}\times\frac{4L}{\pi d^2}=9.7\times10^{-8}\times\frac{4L}{\pi d'^2}\\\\d : d' = 0.42[/tex]

what units of measurement measures both velocity and speed

Answers

Answer:

[tex]metre \: per \: second[/tex]

Explanation:

Velocity is a derived quantity and the S.I unit is metre per second.Speed is also a derived quantity which is has the S.I unit to be metre per second.

Given that o.2i+bj+o.4k is a unit vector,what is the value of b?

Answers

Answer:

b = 0.89

Explanation:

The given vector is, [tex]A=0.2i+bj+0.4k[/tex]

A is a unit vector

We need to find the value of b.

For a unit vector, |A| = 1

So,

[tex]0.2^2+b^2+0.4^2=1\\\\0.04+b^2+0.16=1\\\\0.2+b^2=1\\\\b^2=1-0.2\\\\b=0.89[/tex]

So, th value of b is 0.89.

An electric field is given by in units of N/C when is in meters. What is the potential difference from point B at (0,7) m to point A at (7,0) m

Answers

Complete Question

An electric field is given by E= (5x, 0, 0) in units of N/C when x is in meters. What is the potential difference VA - VB from point B at (0,7) m to point A at (7,0) m? V

Answer:

[tex]V_A-V_B=-122.5V[/tex]

Explanation:

From the question we are told that:

Position of point [tex]B=(0,7) m[/tex]

Position of point [tex]A=(7,0) m[/tex]

Generally the equation for pd across the points is mathematically given by

 [tex]V_A-V_B=-\int_0^7(5x)d[/tex]

 [tex]V_A-V_B=-[\frac{5x^2}{2}]_0^7[/tex]

 [tex]V_A-V_B=-[\frac{5(7)^2}{2}][/tex]

 [tex]V_A-V_B=-122.5V[/tex]

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp

Answers

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm  

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]

A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water

Answers

Answer:

The speed of the swimmer in stil water is 0.5 m/s

Explanation:

Given;

total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s

width of the river, = 200 m

Please find the image attached for explanation.

What is the friction force on a box that has a mass of 15kg as it slides across the floor. The coefficient of friction of the not very clean floor is 0.25
please explain everything including formula used ​

Answers

Answer:

36.75 N

Explanation:

Applying

F = mgμ................. Equation 1

Where F = Friction force on the box, m = mass of the box, g = acceleration due to gravity of the box, μ = coefficient of static friction

From the question,

Given: m = 15 kg, μ = 0.25

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 15(9.8)(0.25)

F = 36.75 N

Hence the friction force on the box is 36.75 N

assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K​

Answers

Answer:

T = 2398 K

Explanation:

To calculate the emission of the light bulb we use the law is Stefan

           P = σ A e T⁴

as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)

           T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]

    let's calculate

           T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]

           T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]

           T = 2,398 10³ K

           T = 2398 K

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s

Answers

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

differentiate between step up and step down transformer​

Answers

Answer:

The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.

you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens

Answers

Answer:

concave curves inward like an hourglass and convex is an outward curve like a football

Explanation:

hope this helps

what is the frequency of a wave related to​

Answers

Answer:

Frequency is the number of complete oscillations or cycles or revolutions made in one second.

You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.

Answers

Answer:

exactly 1273 N whether or not the box moves.

Explanation:

In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction

Therefore as per the given situation, the above represent the answer

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ​

Answers

Answer:

144 meters

Explanation:

it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .

so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144

An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?

Answers

Answer:

450792 kgm/s

Explanation:

by conservation of momentum,

total momentum AFTER collision = total momentum BEFORE collision

                                                       =mv+m'v'

                                                       =12954×28+17616×5

                                                       =450792 kgm/s

A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices

Answers

Answer:

Points away from the disk along the z-axis.

Explanation:

Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.

Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.

Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous

Answers

aluminum has the largest atomic radius

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

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