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Dominik [7]
2 years ago
12

Iron man crashes into a parked car with a force of 8,000 newtons, resulting in a acceleration of 1.14 m/s. if the car has a mass

of 1000 kg what is the coefficient of kinetic friction between the car tires and the road?
Physics
1 answer:
Anna11 [10]2 years ago
8 0

Answer:

0.8

Explanation:

F=uR

F=8,000N

R= 1000*10=10000

u=F/R

u= 8000/10000

u=0.8

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MrRa [10]

Answer:

10m/s

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Newton's _____ law explains why my hands hurt when I clap loudly
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Newtons third law (inertia) is to blame
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A cart travels down a ramp at an average speed of 5.00 centimeters/second. What is the speed of the cart in miles/hour? (Remembe
viva [34]
Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.

( \frac{ 5 cm}{1 sec} ) *( \frac{1m }{100cm})*( \frac{1km}{1000m} )*( \frac{1mi}{1.6km} ) = 0.00003125 mi/sec \\  \\ ( \frac{0.00003125 mi}{1sec} )*( \frac{60sec}{1min} )*( \frac{60min}{1hr} )=0.1125 mi/hr
7 0
3 years ago
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Please help
Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

F = ma

<em>replacing the variables</em>

300 = 10 * a

a = 30 m/s²

5 0
3 years ago
Physics Homework MathPhys homie if you see this pls help
cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0
3 years ago
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