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3 years ago

Work occurs when

Physics

2 answers:

azamat3 years ago

8 0

Answer:

The answer is A. an applied force results in movement of an object in the same direction as the applied force.

Explanation:

Work is the product of a force applied to a body and the displacement of the body in the direction of this force, that is, a force performs a work when there is a displacement of its point of application in the direction of that force. The work of the force on that body will be equivalent to the energy needed to move it.

neonofarm [45]3 years ago

6 0

Work occurs when an applied force results in movement of an object in the same direction as the applied force.

Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work.

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Metals in group 1 on the periodic table most commonly form which type of ion

givi [52]

Answer:

Group 1 - the alkali metals. The Group 1 elements in the periodic table are known as the alkali metals.

Explanation:

7 0

3 years ago

Reducing, reusing, and recycling in your office is likely to _______.

iris [78.8K]

Conserve natural resources, energy and landfill space.

4 0

4 years ago

A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p

a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s$

4 0

3 years ago

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

3 years ago

Which of the following examples could this free body diagram describe? (Check all that apply)

boyakko [2]

Answer:

A car accelerating to the right

Explanation:

The free-body diagram shows all the forces acting on an object. The length of each arrow is proportional to the magnitude of the force represented by that arrow.

In this free-body diagram, we see that there are 4 forces acting on the object, in 4 different directions. We also see that the two vertical forces are equal so they are balanced, while the force to the rigth is larger than the force to the left: this means that there is a net force to the right, so the object is accelerating to the right.

Therefore, the correct answer is:

A car accelerating to the right

5 0

3 years ago