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Amiraneli [1.4K]
3 years ago
5

A 1500 kg car rounds a horizontal curve with a radius of 52 m at a speed of 12 m/s. what minimum coefficient of friction must ex

ist between the road and tires to prevent the car from slipping?
Physics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

The minimum coefficient of friction must exist between the road and tires to prevent the car from slipping is μ= 0.28

Explanation:

m= 1500 kg

r= 52m

Vt= 12m/s

g= 9.8 m/s²

Vt= ω * r

ω= Vt/r

ω= 0.23 rad/s

ac= ω²* r

ac= 2.77 m/s²

Fr= m* ac

Fr= 4153.84 N

W= m*g

W= 14700 N

Fr= μ * W

μ= Fr/W

μ=0.28

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A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?
Anon25 [30]

Answer:

57N

Explanation:

F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }

6 0
3 years ago
Read 2 more answers
Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m
STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]

7 0
3 years ago
A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro
Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

v=\omega \times r

1.25\times 100=\omega \times 3.7

\omega =\frac{125}{3.7}=33.78

and \frac{2\pi N}{60}=\omega

N=\frac{\omega \times 60}{2\pi }

N=\frac{33.78\times 60}{2\pi }

N=322.53 rpm

8 0
3 years ago
ILL GIVE BRAINLY THING
Bogdan [553]

Answer:

About 3 trips

Explanation: if we do 2.5m*1.6m*0.75 it equals to 11000 then we divide that to 11m3 and it gives you 3.6 so it will be about 3 times

Thx

4 0
4 years ago
Question 1 (1 point)
Juli2301 [7.4K]

Hello!

\large\boxed{800Ns}

Remember that impulse is equivalent to:

Impulse = force (N) × time (s)

Plug in the given force and time:

Impulse = 25 × 32

Impulse = 800 Ns

3 0
3 years ago
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