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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 sound wave has a frequency of 250 Hz and a wavelength of 2.5 m. What is the speed of the wave?
Digiron [165]

ANSWER

625 m/s

EXPLANATION

Given:

• The frequency of the sound wave, f = 250 Hz

,

• The wavelength, λ = 2.5 m

Find:

• The speed of the wave, v

The speed of a wave of wavelength λ and frequency f is given by,

v=f\cdot\lambda

Substitute the known values and solve,

v=250Hz\cdot2.5m=625m/s

Hence, the speed of the wave is 625 m/s.

3 0
1 year ago
a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
GalinKa [24]

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

∴acceleration, a= 14/70 = 0.2 m/s²

now on applying conservation of linear momentum

pi=pf            pi=0 (initially at rest)

0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s

v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

0- (0.3) ²= -2 x 0.2 x 5

s= 0.09/0.4 ≈ 0.3 m

Learn more about distance here-

brainly.com/question/15172156

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6 0
1 year ago
Need Help With Physical Ed
Dimas [21]

which of the following is not a barrier to physical activity it is fear of injury I think.

8 0
3 years ago
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( Image is 6 C carbon with the numbers 12.011 under it ) According to the image, the atomic mass of carbon is 12.011. How is the
Oduvanchick [21]

Answer:

B. By adding the number of protons and the number of neutrons

Explanation:

The atomic mass is determined by adding the number of protons and neutrons in an atom. An atom is made up of three fundamental particles: Electrons, Protons and Neutrons.

The protons and neutrons occupy a central region in an atom known as the nucleus. The nucleus is positively charged and mass concentrated.

If we compare the relative masses of the subatomic particles, the masses of protons and neutrons would be 1 and that of an electron would be 1/1840. This shows that the mass of electrons are negligible.

In order to ascertain atomic mass, we therefore add the number of protons and neutrons together. This is how we arrive at 12.011 as the value of the atomic mass of C and for other elements.

The atomic mass is also known as the mass number.

4 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
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