All categories

3 years ago

50 J of work was performed in 20 seconds. How much power was used to perform this task? A. 0.4 W B. 2.5 W C. 4 W D. 24.5 W

Physics

1 answer:

Ugo [173]3 years ago

5 0

Answer:

B. 2.5 W

Explanation:

Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).

Mathematically,

Power = Energy or work/Time

P = W/t ........................Equation 1

Where P = power used to perform the task, W = work, t = time.

Given: W = 50 J, t = 20 s.

Substitute into equation 1

P = 50/20

P = 2.5 W.

Hence the power is 2.5 W.

The right option is B. 2.5 W

You might be interested in

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

-10 + 12 sin 60 + Fₓ = 0

Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

12 cos 60 - 6 + F_y = 0

F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N

Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

$F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}$

F = 0.39N

We use trigonometry for the angle.

$tan \theta = \frac{F_y}{F_x}$

tan θ= 0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

θ = 180 + θ'

θ = 180 + 0

θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

10+ 6 + 6 + F = 0

F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

$a = \frac{F}{m}$

a = $\frac{24}{6}$

a = 4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

v = v₀ - a t

Final velocity when stopped is zero

t = $\frac{0-v_o}{a}$

t = 100/4

t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

2 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-

kifflom [539]

Magnitude of the force of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

The force of tension, $F_T$ , acting up along the plane
The force of friction, $F_f=14.8 N$ , acting down along the plane
The component of the weight in the x-direction, $F_{gx}$ , acting down along the plane