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2 years ago

12

What happens to the orbit time as the distance increases.

1 answer:

Maslowich2 years ago

6 0

The orbital period increases if the orbital distance is increased.

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

2 years ago

Imagine imagining an imagination.

ozzi

I’m imagining imagining imagining an imagination...

4 0

3 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

now here we can say

$m_1 = 10 g$

$v_{1i} = 0$

$m_2 = 550 g$

$v_{2i} = 3.5 m/s$

now here we can say

$10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}$

$192.5 = v_{1f} + 55 v_{2f}$

now by coefficient of restitution

for elastic collision we know that e = 1

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

$v_{2f} - v_{1f} = 0 - 3.5$

now by solving the two equation

$56v_{2f} = 189$

$v_{2f} = 3.375 m/s$

also we know that

$v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s$

so final speed of the nail is 6.875 m/s

6 0

2 years ago

Read 2 more answers

Where can you find the neutrons of an atom? in the nucleus with the protons orbiting the nucleus In the nucleus with the electro

cricket20 [7]

Answer:

Nuclease is the answer I know

I hope this is the answer

3 0

3 years ago

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc

Nuetrik [128]

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

$v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2$

The data given are;

m1 = 281kg
u1 = 2.82m/s
m2 = 209kg
u2 = -1.72m/s
v1 = ?

Let's substitute the values into the equation.

$v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s$

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

4 0

2 years ago