PHYSICS COLLEGE

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.

Answers

Answer 1

Answer:

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^(\circ)$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50* 9* \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

Related Questions

A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.Calculate di/dtdi/dt = _________.
A system of 1223 particles, each of which is either an electron or a proton, has a net charge of -5.328×10-17 C. How many protons are in this system (exactly)?
A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.
Analysis of the relationship between the fuel economy (mpg) and engine size (liters) for 35 models of cars produces the regression model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter engine, what does this model suggest the gas mileage would be?
Imagine that you drop an object of 10 kg, how much will be the acceleration andhow much force causes the acceleration?

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?

Answers

Answer:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

$mgx - (1)/(2)kx^2 = 0$

$(2.07)(9.8)(0.131) - (1)/(2)k(0.131)^2 = 0$

now we will have

$k = 310 N/m$

Part B)

Time period of oscillation is given as

$T = 2\pi\sqrt{(m)/(k)}$

$T = 2\pi\sqrt{(2.07)/(310)}$

$T = 0.51 s$

A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?

Answers

To solve this problem it is necessary to apply the related concepts to the moment of inertia in a disk, the conservation of angular momentum and the kinematic energy equations for rotational movement.

PART A) By definition we know that the moment of inertia of a disk is given by the equation

$I = (1)/(2) MR^2$

Where

M = Mass of the disk

R = Radius

Replacing with our values we have

$I = (1)/(2) (0.2)(0.3)^2$

$I = 9*10^(-3)kg\cdot m^2$

The initial angular momentum then will be given as

$I = I_1 \omega_1$

$I = 0.012*2$

$I = 0.024kg\cdot m^2/s$

Therefore the total moment of inertia of the table and the disc will be

$I_2 = 9*10^(-3)+0.012$

$I_2 = 0.021kg\cdot m^2$

The angular velocity at the end point will be given through the conservation of the angular momentum for which it is understood that the proportion of inertia and angular velocity must be preserved. So

$I_1 \omega_1 = I_2\omega_2$

$(0.012)(2)=(1.08*10^(-4))\omega_2$

$\omega_2 = (0.012*2)/(0.021)$

$\omega_2 = 1.15rad/s$

Therefore the new angular velocity is 1.15rad/s

PART B) Through the conservation of rotational kinetic energy we can identify that its total change is subject to

$\Delta KE = (1)/(2)I_1\omega_1^2-(1)/(2)I_2\omega^2$

$\Delta KE = (1)/(2)(I_1\omega_1^2-I_2\omega^2)$

$\Delta KE = (1)/(2)(0.024*2^2-0.021*1.15^2)$

$\Delta KE = 0.034J$

Therefore the change in kinetic energy is 0.034J

While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of increasing photon energy, lowest first.

Answers

Answer:

Explanation:

1. radio waves from am

2. radio waves from fm

3.yellow light from a sodium street lamp

4. microwaves from an antenna of a communications system.

2H is a loosely bound isotope of hydrogen, called deuterium or heavy hydrogen. It is stable but relatively rare — it form only 0.015% of natural hydrogen. Note that deuterium has Z = N, which should tend to make it more tightly bound, but both are odd numbers.Required:
Calculate BE/A, the binding energy per nucleon, for 2H in megaelecton volts per nucleon

Answers

Answer:

0.88 MeV/nucleon

Explanation:

The binding energy (B) per nucleon of deuterium can be calculated using the following equation:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u$

Where:

Z: is the number of protons = 1

N: is the number of neutrons = 1

$m_(p)$ : is the proton's mass = 1.00730 u

$m_(n)$ : is the neutron's mass = 1.00869 u

M: is the nucleu's mass = 2.01410

A = Z + N = 1 + 1 = 2

Now, the binding energy per nucleon for ²H is:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u = (1*1.00730 + 1*1.00869 - 2.01410)/(2)*931.49 MeV/u = 9.45 \cdot 10^(-4) u*931.49 MeV/u = 0.88 MeV/nucleon$

Therefore, the binding energy per nucleon for ²H is 0.88 MeV/nucleon.

I hope it helps you!

Final answer:

The binding energy per nucleon for 2H (deuterium) is 1.1125 MeV per nucleon.

Explanation:

The binding energy per nucleon, or BE/A, can be calculated by dividing the total binding energy of the nucleus by the number of nucleons. To calculate the BE/A for 2H (deuterium), we need to know the total binding energy and the number of nucleons in deuterium. The total binding energy of deuterium is approximately 2.225 MeV (megaelectron volts) and the number of nucleons is 2. Therefore, the BE/A for 2H is 2.225 MeV / 2 = 1.1125 MeV per nucleon.

Learn more about Binding energy per nucleon here:

brainly.com/question/10095561

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A certain lightning bolt moves 40 C of charge. How many fundamental units of charge |qe| is this?

Answers

Answer to A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge | qe | is this? . ... charge, N is the total number of electron or protons that constitute total charge Q.

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

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Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of the square is 2a. What is the electric potential at P, the center of the square?a. kQ/4a b. kQ/a c. zero volts d. 2kQ/a e. 4kQ /a