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vesna_86 [32]
3 years ago
8

One hazard of space travel is debris left by previous missions. there are several thousand objects orbiting earth that are large

enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. calculate the force exerted in newtons by a 0.160 mg chip of paint that strikes a spacecraft window at a relative speed of 3.00 ✕ 103 m/s and sticks, given the collision lasts 6.00 ✕ 10−8 s. (enter the magnitude.)
Physics
1 answer:
iren2701 [21]3 years ago
4 0
The chip, during the collision, has a change in momentum:
Δp = m · (v₂ - v₁)

The final speed is equal to zero, since it sticks to the spacecraft, therefore:
Δp = m · v
      = 1.6×10⁻⁷ · 3×10³
      = 4.8×10⁻⁴ kg·m/s
where we transformed the mass into the proper units of measurement (kg).

This change in momentum is equal to the impulse J:
Δp = J = F · t

We can solve for F
F = J / t = <span>Δp / t
   = </span>4.8×10⁻⁴ / 6×10⁻⁸
   = 8.0×10³ N

Hence, <span>the force exerted by the chip on the spacecraft is F = 8000N.</span>
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A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
A set of crash tests consists of running a test car moving at a speed of 10.2 m/s (22.8 mi/hr) into a solid wall. Strapped secur
Len [333]

Answer: The average force is 3934.71N

Explanation: Please see the attachments below

6 0
3 years ago
Động cơ của một ô tô thực hiện một lực kéo không đổi là F=4000N . Biết ô tô chuyển động đều với vận tốc 36km/h . Trong năm phutd
erik [133]

Answer:

The work done is 12 MJ.

Explanation:

The engine of a car exerts a constant traction force of F=4000N . Assume that the car is moving at a constant speed of 36 km/hr. In five minutes, what is the work done by the engine's traction?

Force, F = 4000 N

speed, v = 36 km/h = 10 m/s

time , t = 5 minutes = 5 x 60 = 300  s

Work done is given by

W =  force x distance

W = 4000 x 10 x 300 J = 12 x 10^6 J = 12 MJ

3 0
3 years ago
In which situation is no work being done? A. a person carrying a box from one place to another B. a person picking up a box from
tatiyna

Answer:

The answer is A

Explanation:

Being the force is acting upwards and displacement takes place horizontally

3 0
3 years ago
One car is sitting still the other is moving at a velocity of 4m/s. If two cars have a mass of 0.04 kg run into each other, what
Leni [432]

Answer:

p = 0.16 kgm/s

Explanation:

the initial momentum combined of the two cars and the final momentum of the paired cares are the same, so we just need to find the initial momentum

p = m1v1 + m2v2

p = 0.04*4 + 0.04*0

p = 0.04*4

p = 0.16 kgm/s

7 0
3 years ago
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