Ask question

Ask question

All categories

2 years ago

10

A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk

ateboarder is m = 82 kg. The coefficient of static friction between the surface of the parking lot and the wheels of the skateboard is μs = 0.63.
What is the maximum speed, in meters per second, he can travel through the arc without slipping?

1 answer:

Sonbull [250]2 years ago

3 0

To solve this problem we will apply the concepts related to the centripetal force, for which it is necessary to equate it with the static friction force of the body. From this, we will clear the speed and replace with the given values. Our values are defined as,

$r = 16m$

$m = 82kg$

$\mu_s = 0.63$

Maximum velocity can be find out using centripetal force,

$F_c = \frac{mv^2}{r}$

Must be equal to,

$\frac{mv^2}{r} = \mu_s mg$

$v = \sqrt{\mu_s gr}$

$v = \sqrt{(0.63)(9.8)(16)}$

$v = 9.93m/s$

Therefore the maximum speed that he can travel through the arc without slipping is 9.93m/s

You might be interested in

A boat is headed with a velocity of 18 meters/second toward the west with respect to the water in a river. If the river is flowi

IrinaK [193]

If the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.

7 0

2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Negle

Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0

2 years ago

Based on the free-body diagram, the net force acting on this wheelbarrow is {blank} N.

Semenov [28]

Answer:

diagram?

Explanation:

5 0

2 years ago

Read 2 more answers

Which of the following best describes an action-reaction pair? A. The Moon Pulls on Earth, and Earth pulls back on the moon. B.

Papessa [141]

An action-reaction pair would be a pair in which one of the elements exerts a force on the other element (action), and then the other element would respond to this force by exerting another force in the opposite direction (reaction).

From the given choices, we will see that:
For choice A, the moon exerts a force on the earth by pulling it (action) and the earth responds to this force by pulling the moon (reaction in opposite direction of the action).

Therefore, the correct choice would be:
A. <span>The Moon Pulls on Earth, and Earth pulls back on the moon.</span>

4 0

2 years ago

Read 2 more answers