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2 years ago

A center-seeking force related to acceleration is _______ force.

1 answer:

4 0

A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.

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What is a non contact force that attracts all objects to the centre of the earth

beks73 [17]

Answer:

Gravity. The weight-force or weight of an object is the force because of Gravity, which acts on the object attracting it towards the centre of the earth.

Hope this helps, 

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4 0

3 years ago

If two opposing forces are equal, then the net force is 0 N. true or false?

Fofino [41]

Answer:

false

Explanation:

6 0

2 years ago

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

Acceleration of the cart:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

7 0

2 years ago

Read 2 more answers

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

What is kinematics ?? xD -,- xD

lorasvet [3.4K]

Answer:

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Explanation:

Hope this helps!

Blazetoxic123

7 0

3 years ago