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inn [45]
3 years ago
11

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

of 1.0 x 10 2 meters per second 2 what is the magnitude of the force the insect exerts on the car
Physics
1 answer:
Mariana [72]3 years ago
6 0
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass m=1.0 \cdot 10^{-4} kg. The acceleration of the insect is a=1.0 \cdot 10^2 m/s^2, therefore we can calculate the force exerted by the car on the insect:
F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
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A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla
MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, u=3\times 10^5\ m/s

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s

So, the speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

5 0
3 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
Select all the correct locations on the image.
Papessa [141]

Answer:

its the top 3 can confirm on plato

Explanation:

8 0
2 years ago
Read 2 more answers
You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am
emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

1 - (v₀/343) = 0.9141

v₀/343 = 1 - 0.9141 = 0.0859

v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour

7 0
3 years ago
Use the drop-down menus to complete the statements. When the 5.0 kg cylinder fell 100 m, the final temperature of the water was
OLEGan [10]

Answer:

A. 26.17 B. 1.17 C. 30.86 D. 5.86

Explanation:

7 0
3 years ago
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