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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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Geological evidence based on several radiometric techniques has provided a scientifically well-accepted age for the Earth. Repre
lidiya [134]
Considering earth is 4.5 billion years old and dinosaurs only went extinct 66 million ago o would say Answer: A - a little over one yard
3 0
2 years ago
Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo
zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

3 0
2 years ago
You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m
mamaluj [8]
The correct answer is:
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7 0
2 years ago
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