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valina [46]
2 years ago
13

While a car is stopped at a traffic light in a storm, raindrops strike the roof of the car. The area of the roof is 5.0 m2. Each

raindrop has a mass of 3.7 ✕ 10−4 kg and speed of 2.5 m/s before impact and is at rest after the impact. If, on average at a given time, 150 raindrops strike each square meter, what is the impulse of the rain striking the car?
Physics
1 answer:
julsineya [31]2 years ago
5 0

Answer:

J = 0.693 N.s

Explanation:

The impulse of one single drop is given by:

J1 = m*(Vf - Vo)   where Vf = 0

J1 = -9.25*10^{-4}N.s

The magnitude of the total impulse will be:

Jt = J1 * 150 * 5

Jt = 0.693 N.s

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
Now assume that the boat is subject to a drag force fd due to water resistance. is the component of the total momentum of the sy
DochEvi [55]
Based on the given details with this question, I can say that the direction of motion is not conserved. This is because the boat is subjected to an external force because of water resistance. So, the answer for this question would be NO.
4 0
1 year ago
Read 2 more answers
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

\Delta L=\dfrac{PL}{AE}

A=\dfrac{PL}{\Delta LE}

A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

4 0
2 years ago
Read 2 more answers
A straight, horizontal length of copper wire has a current i=28A through it. What are the magnitude and direction of the minimum
seropon [69]

Answer:

0.01631 T

Explanation:

current, i = 28 A

mass per unit length, m/l = 46.6 g/m = 0.0466 kg/m

Let the magnetic field is B.

the weight of the wire is balanced by the magnetic force .

mg = i l B

B = mg / i l

B = (m/l) x g/i

B = 0.0466 x 9.8 / 28

B = 0.01631 T

Thus, the magnetic field is 0.01631 T.

6 0
2 years ago
A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel
schepotkina [342]
20.3 divided by 3.0 will get u velocity and v times 3.0s 
3 0
2 years ago
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