Answer:

**Answer:**

(a):

(b):

(c):

(d):

(e):

**Explanation:**

**Given, **the position of the particle along the x axis is

The units of terms and should also be same as that of** x**, i.e.,** meters**.

The unit of** t** is** seconds**.

**(a):**

Unit of

Therefore, unit of

**(b):**

Unit of

Therefore, unit of

**(c):**

The velocity **v** and the position **x** of a particle are related as

**(d):**

The acceleration** a **and the velocity **v** of the particle is related as

**(e):**

The particle attains maximum **x **at, let's say, , **when the following two conditions are fulfilled**:

Applying both these conditions,

For ,

Since, c is a positive constant therefore, for ,

Thus, particle does not reach its maximum value at

For ,

Here,

Thus, **the particle reach its maximum x value at time **

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?Express your answer in terms of the variable v0 and appropriate constants.c) For what value of h does the collision occur at the instant when the first ball is at its highest point?Express your answer in terms of the variable v0 and appropriate constants.

A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!Please help ASAP

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

A rocket accelerates upwards at 6.20 ft/s/s. How far will the rocket travel in 2 minutes?

Which lists mixtures, in order, from the smallest particles to the largest particles?O solution, colloid, suspensionsolution, suspension, colloidO suspension, colloid, solutionO suspension, solution, colloid

A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!Please help ASAP

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

A rocket accelerates upwards at 6.20 ft/s/s. How far will the rocket travel in 2 minutes?

Which lists mixtures, in order, from the smallest particles to the largest particles?O solution, colloid, suspensionsolution, suspension, colloidO suspension, colloid, solutionO suspension, solution, colloid

**Answer:**

**a) greatest voltage = 29.25 V**

**b) power = 16 W**

**Explanation:**

The total resistance R of the three resistors in series is:

** **

a) The greatest current **I** is the one that will burn the resistor with lower power rating, which is 9.12 W:

The voltage is:

b) When the current is 0.54 A, the power is:

The **work** () done by the friction force between the **ramp** and the skateboarder is given by .

The **work****done** by the friction force () can be calculated using the formula for work, which is the product of the force applied () and the displacement (d) over which the **force** is applied:

In this scenario, the **friction****force** works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of **skateboarder** () = 54 kg

Height of the ramp () = 3.3 m

Final velocity () = 6.2 m/s

Coefficient of **kinetic****friction** () between skateboarder and ramp

Acceleration due to gravity (g) =

The normal **force** () is equal to the weight of the skateboarder:

The **displacement** (d) is the vertical distance () that the **skateboarder** descends down the ramp.

Now we can write the **expression** for the work done by the friction force ():

Substitute the **expression** for the normal force:

Thus, this expression represents the work done by the **friction force** between the ramp and the **skateboarder** in terms of the given variables.

For more details regarding **friction force** visit:

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Your question seems incomplete, the probable complete question is:

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

The **momentum **of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

To find the momentum of the box, we can use the formula:

**Momentum = mass x velocity**

**a**. The momentum of the box with respect to the floor is: 5 kg x 5 m/s =** 25 kg·m/s**

**b**. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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time = energy / power = 15010 / 500 = .... seconds

**Answer:**

**6.5e-4 m**

**Explanation:**

We need to solve this question using law of conservation of energy

**Energy at the bottom of the incline= energy at the point where the block will stop**

Therefore,** Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=**

**Energy at the point where the block will stop consists of only gravitational potential energy=**

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒

⇒

Also

where is the mass of block

is acceleration due to gravity=9.8 m/s

is the difference in height between two positions

⇒

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴

⇒

⇒

**⇒h=0.0006467m=**

**Answer:**

E = 1.76 J

**Explanation:**

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

So, the change in its gravitational potential energy is 1.76 J.

This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.

This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.

To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.

Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:

0.35*F2 = 3.25*F1 (1)

Because the net force should also be zero in equilibrium, we have:

F1 + F2 = 205.8 (2)

Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.

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To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a **force** of approximately 114N is required

The subject question is a classic example of **Torque** problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The **weight** of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.

This torque is calculated as Torque = r * F = 1.8m (**distance** from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.

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