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4 years ago

Which term s the product of force and distance?

Physics

1 answer:

goldfiish [28.3K]4 years ago

8 0

Answer:

The answer should be WORK 

(Hope this Helps)

Explanation:

work is equal to the product of force and distance.

W=F*S

W=Work

F=Force applied

S=Distance

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1. How long would the car in Sample Problem C take to come to a stop from its initial velocity of 20.0 m/s to the west? How far

Brilliant_brown [7]

The time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.

<h3>What is the distance?</h3>

The length of the path traveled by the body is known as the distance covered by the body.

Distance is a 1-dimensional phenomenon. its unit is a meter(m). The distance can be found by the product of velocity and time.

The given data in the problem will be

u is the initial velocity =20m/sec

t is the time =?

d is the distance =?

From the Newtons second law;

$\rm F \triangle t = \triangle P \\\\ \triangle t = \frac{\triangle P }{F} \\\\ \triangle t = \frac{m(v_f-v_i)}{F} \\\\ \ \triangle t = \frac{2240 (0-20))}{8410} \\\\ \triangle t = 5.3 \ sec \\\$

The distance travelled before the car stop is,

$\rm \triangle t = \frac{1}{2} (v_f+v_i)\traingle t \\\\ \traingle x = \frac{1}{2} (-20+0)5.3 \\\\\ \triangle x= -53.3 m \ west$

Hence,the time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.

To learn more about the distance, refer to the link;

brainly.com/question/989117

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2 years ago

A 645-turn coil with a 20.250 m 2 area is spun in Earth’s 5.00×10 −5 T magnetic field, producing a 1.25-V maximum emf. A

Dmitriy789 [7]

Answer:

$\omega = 1.914\ rad/s$

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

$\varepsilon = NAB\omega$

$\omega= \dfrac{\varepsilon}{NAB}$

$\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}$

$\omega = 1.914\ rad/s$

Angular velocity of the coil = $\omega = 1.914\ rad/s$

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4 years ago

Scientists are studying a group of living cells under a microscope. They write down notes about the size, shape, and color of th

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Answer:

Explanation:

descriptive, because scientists are writing down the observations but not making comparisons.

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3 years ago

A ball of mass 1.6 kg is attached to the end of a massless string. A circus clown twirls the

erik [133]

Compute the ball's angular speed v :

v = (1 rev) / (2.3 s) • (2π • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s

Use this to find the magnitude of the radial acceleration a :

a = v ²/R

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a = v ² / (180 cm) = v ² / (1.8 m) ≈ 13.43 m/s²

The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude

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3 years ago

An astronaut on a strange new planet finds that she can jump up to a height of 72 m before she would fall back down to the surfa

gregori [183]

Answer:

25m/s²

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v is the final velocity of the astronaut

u is his initial velocity

a = -g (the acceleration will be acceleration due to gravity since he is acting under the influence of gravity. The value is negative because the astronaut jumps up to a particular height)

s = H = total height covered

The equation will then become;

v² = u²-2gH

Given

u = 60m/s

v = 0m/s

g = ?

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Substituting the given value into the equation;

0² = 60²-2g(72)

0 = 3600-144g

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g = -3600/-144

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The magnitude of his acceleration due to gravity on the planet is 25m/s²

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3 years ago