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finlep [7]
3 years ago
13

Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to

Physics
1 answer:
fenix001 [56]3 years ago
6 0

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

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The acceleration of a cart rolling down a ramp depends on __________.
zmey [24]

The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.

8 0
3 years ago
A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc
Stells [14]

Answer:

<em>The centripetal acceleration would increase by a factor of 4</em>

<em>Correct choice: B.</em>

Explanation:

<u>Circular Motion</u>

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

v_t=w\ r

\displaystyle a_c=\frac{v_t^2}{r}

If the speed is doubled and the radius is the same, then

\displaystyle a_c=\frac{(2v_t)^2}{r}

\displaystyle a_c=4\frac{v_t^2}{r}

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

5 0
3 years ago
A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?​
Rainbow [258]

Answer: 113.75

Explanation:

You know

acceleration = a = 3.5 m/s²

time = t = 5 seconds

initial velocity = u = 14 m/s

Unknown is distance = s = ?

Use equation: s = ut + \frac{1}{2} at²

Substitute all the known values inside the equation:

s = (14*5) + 0.5 * 3.5 * 5²

s = 70 + 43.75 = 113.75 m

The car travels 113.75 metres.

3 0
3 years ago
While Newton's Second Law often deals with formulas and numerical calculations, there exist one very important nonnumerical conv
zheka24 [161]

Option B is the correct answer.

MKS system gives the following units:

Distance ----- meters

Mass ----- Kilograms

Time ----- seconds

meter is basic unit for length measurement. smaller units are centimeter, millimeter, micrometer, bigger units are kilometer and so on.

kilogram is the basic unit for mass. smaller unit is gram.

second is the basic unit for time. Greater units are minutes, hours, smallest unit are micro second and so on.

8 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
Olin [163]
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law). 

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: \mu m g, where \mu=0.25 is the coefficient of friction and m=30.0 kg is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N
And so, this is the force that the worker must apply to the crate.
5 0
3 years ago
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