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Phantasy [73]
3 years ago
5

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

ring

Physics
2 answers:
AnnZ [28]3 years ago
5 0
Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

Sedaia [141]3 years ago
4 0

<u>Answer:</u> The tension in the string is 6 N

<u>Explanation:</u>

Tension is defined as the pulling force that is applied on a string, chain etc.. Force is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=ma

where,

F = tension or force exerted in the string

m = mass of the cart pulling the string = 3.00 kg

a = acceleration of the string = 2.00m/s^2

Putting values in above equation, we get:

F=3.00kg\times 2.00m/s^2\\\\F=6N

Hence, the tension in the string is 6 N

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7 0
3 years ago
What is an essential characteristic of an object in equilibrium?
bulgar [2K]

Answer:

2) zero acceleration

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

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In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

Generally, an object is said to be in equilibrium when neither the energy possessed by the object not state of motion changes with respect to time. Thus, the vector sum of all the forces acting upon an object that's in equilibrium is zero.

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5 0
3 years ago
A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into
vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2}  + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s

5 0
3 years ago
Read 2 more answers
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a
timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}

\omega \approx 17.781\,\frac{rad}{s}

Then, the tangential speed of the tack is: (\omega \approx 17.781\,\frac{rad}{s}, R = 0.393\,m)

v = \left(17.781\,\frac{rad}{s} \right)\cdot (0.393\,m)

v = 6.988\,\frac{m}{s}

The tangential speed of the tack is 6.988 meters per second.

7 0
3 years ago
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scoundrel [369]
The answer would be b. cleavage 
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