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3 years ago

An object needs a force of 152 Newtrons to move 8 meters. How much work is required?

Physics

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

960J

Explanation:

Given parameters:

Force = 120N

Distance = 8m

Unknown:

Work required = ?

Solution:

The work done by a body is the force applied to move a body in a specific direction.

Work done = Force x distance

Insert the parameters and solve;

Work done = 120 x 8 = 960J

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Acceleration is often measured in what unit? A. newtons B. kilograms C. meters per second D. meters per second squared

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D. meters per second squared

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3 years ago

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3 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0

3 years ago

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0

3 years ago