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Masja [62]
3 years ago
14

A cart with an unknown mass is at rest on one side of a track. A student must find the mass of the cart by using Newton’s second

law. The student attaches a force probe to the cart and pulls it while keeping the force constant. A motion detector rests on the opposite end of the track to record the acceleration of the cart as it is pulled. The student uses the measured force and acceleration values and determines that the cart’s mass is 0.4kg . When placed on a balance, the cart’s mass is found to be 0.5kg . Which of the following could explain the difference in mass?
Answer choices:

A) The track was not level and was tilted slightly downward.

B) The student did not pull the cart with a force parallel to the track.

C) The wheels contain bearings that were rough and caused a significant amount of friction.

D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.
Physics
1 answer:
konstantin123 [22]3 years ago
5 0

Explanation:

D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.

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two electronics students are discussing electrical units. student A says that electrical power is measured in units coulombs, st
Savatey [412]
The statements of both students are incorrect.

-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms'  means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).

There are no 'following statements'.

All in all, a very disappointing question.

8 0
3 years ago
An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object
EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

5 0
3 years ago
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
The tongue weight of a trailer should be what percent of the gross trailer weight rating
mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

-  Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0
3 years ago
A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en
dem82 [27]

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

7 0
4 years ago
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