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Tresset [83]
1 year ago
5

A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

etic friction between the crate and the floor?
Physics
1 answer:
____ [38]1 year ago
6 0

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

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During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
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1.A) 4.9 m  

AL2006 Ace

The instant it was dropped, the ball had zero speed.


After falling for 1 second, its speed was 9.8 m/s straight down (gravity).


Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.


Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.


ANYTHING you drop does that, if air resistance doesn't hold it back.


Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

5 0
3 years ago
Read 2 more answers
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially
mixer [17]

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0
2 years ago
My teacher didnt explain this well and im lost
Vedmedyk [2.9K]

Explanation:

(a) The given figure is a convex lens.

(b) In this figure, the object is placed between F and optical center of a lens. Convex lens is a converging lens. It converges the beam of light falling on it after reflection. The image is formed on the same side of the lens as the object.

The formed image is enlarged and it is virtual and erect.

(i) Type : virtual

(ii) Orientation : upright

(iii) Size : Enlarged

4 0
3 years ago
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.
Black_prince [1.1K]
<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.
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