All categories

3 years ago

State the term using to describe the turning force force exerted by the man

Physics

1 answer:

defon3 years ago

4 0

Answer:

The distance between the line of action of force and the axis of rotation (or pivoted point)

Explanation:

The distance between the line of action of force and the axis of rotation (or pivoted point) .

You might be interested in

Will mark brainliest

sweet-ann [11.9K]

Answer: Wouldn't it just be her blocks all walked in an hour added together?

Explanation: 5+2+3+2=12 so 12 blocks an hour?

3 0

3 years ago

Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1450 N . Assume that the pla

juin [17]

To solve this problem we will apply the concept of Impulse. Which is described as the product between the Force and the change in time. Mathematically this can be described as

$I = F \Delta t$

Where,

F = Force

$\Delta t$ = Time

Our values are given as,

F = 1450N

$\Delta t = 5.1*10^{-3} s$

Replacing we have,

$I = (1450)(5.1*10^{-3})$

$I = 7.395Kg\cdot m/s$

Therefore the impulse delivered to the soccer ball is $7.395Kg\cdot m/s$ or $7.395N\cdot s$

4 0

3 years ago

A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction

Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed = 25.5m/s

Acceleration = 1.94m/s²

Time = 2.3s

Unknown:

Final speed of the car = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

Now insert the parameters and solve;

v = 25.5 + (1.94 x 2.3) = 29.96m/s

3 0

3 years ago

When submerged in water, the brick appears to have an additional force pushing it ___upward ___downward. This additional force i

Neko [114]

Answer:

Therefore, the brick appears to have an additional force pushing it upward.

Explanation:

When a brick is submerged in the water, it has two forces acting upon it. One force is the gravitational force or the weight of the brick, that acts downward. The weight force also acts on the brick when it is not in water. But, in water an additional force acts on the brick. This additional force is named as Buoyant Force. This force is equal to the weight of the water displaced by the brick. And this Buoyant Force acts on the brick in the upward direction. The formula for this force is given as follows:

Buoyant Force = (Density of Water)(Volume of Water Displaced)(g)

<u>Therefore, the brick appears to have an additional force pushing it upward.</u>

<u></u>

8 0

4 years ago

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because

Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

$\tau = F*d$

Where

F= Force

d = Distance

Our values are given as,

$m_1 = 0.32Kg$

$m_2 = 0.4Kg$

$d = 8.7*10^{-2}m$

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

$\tau = T_2-T_1$

$\tau = F_2*d-F_1*d$

$\tau = m_2g*d-m_1*g*d$

$\tau = (m_2-m_1)g*d$

$\tau = (0.4-0.32)(9.8)(8.7*10^{-2})$

$\tau = 0.068N\cdot m$

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is $\tau = 0.068N\cdot m$

8 0

3 years ago