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Flura [38]
3 years ago
13

To test the resilience of its bumper during low-speed collisions, a 4 060-kg automobile is driven into a brick wall. the car's b

umper behaves like a spring with a force constant 8.00 106 n/m and compresses 3.72 cm as the car is brought to rest. what was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Simora [160]3 years ago
4 0
Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
Spring compression, x = 3.72 cm = 3.72 x 10⁻² m

Let the car strike the wall with speed v m/s.

The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
(1/2)mv² = (1/2)kx²
(4060 kg)*(v m/s)² = (8 x 10⁶ N/m)(3.72 x 10⁻² m)²
4060 v² = 1.1701 x 10⁴
v = 1.6513 m/s

Answer:  1.65 m/s (nearest hundredth)

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Answer:

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What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7
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Which statement describes characteristics of a concave lens?
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5 0
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Read 2 more answers
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
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