All categories

2 years ago

It takes 500W of power to move an object 96 m in 12 s. What force is being applied to the object?

Physics

1 answer:

Marianna [84]2 years ago

5 0

The force being applied to the object is calculated to be 62.5N.

<h3>POWER:</h3>

Power is the rate at which work is done. It is measured in watts (W). Power can be calculated by multiplying the force applied on a given object by its speed.

Power (P) = Force (F) × speed (V)

According to this question, it takes 500W of power to move an object 96 m in 12 s. The speed of the object in m/s is calculated as: 96m/12s = 8m/s.

The force applied on the object = 500/8

Force = 62.5N

Therefore, the force being applied to the object is calculated to be 62.5N.

Learn more about force at: brainly.com/question/1436187

You might be interested in

The universe could be considered an isolated system because

Stolb23 [73]

Answer:

A(many people think that no energy or matter exists outside the universe)

Explanation:

4 0

3 years ago

Can someone help me asap this is due at 2 50

Rom4ik [11]

The data is inadequate

8 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Groups of students are making measurements of small cars rolling across a table. the class has measuring tapes with markings in

TiliK225 [7]

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Centimeters Measurements:

A centimeter is a metric unit of measurement used for measuring the length of an object, It is written as cm
Centimeter is one hundredth of a meter, 1 meter is 0.01 cm.

Inches measurements:

An inch can be defined as a unit of length in the customary system of measurement. Length in inches is either represented by in or ''.
1 meter is equal to 39.37 inches

here, the cars are small objects.

The number of centimeters is always bigger,

because a centimeter unit is smaller than an inch unit, and it takes more of them when we are measuring.

Hence,

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Learn more about accurate measurement here:

<u>brainly.com/question/4119127</u>

<u />

#SPJ4

4 0

2 years ago

Which of the following would decrease the resistance in a wire?

padilas [110]

<u>Increase the thickness of the wire</u> would decrease the resistance in a wire

Explanation:

Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.

Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.

6 0

3 years ago