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3 years ago

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A force does 210 J of work when it acts on a moving object and its direction is in the same direction as the object’s displaceme

nt. How much work does this force do when the angle between it and the object’s displacement is 56°

1 answer:

MariettaO [177]3 years ago

5 0

Answer:

When the angle is 56° the work done is 117.43 J

Explanation:

Work = F . s = Fscosθ

We have

W1 = 210 J, θ = 0°

Substituting

210 = F x s x cos 0 = Fs

Now we have to find W2 when angle θ = 56°

Substituting

W2 = F x s x cos 56 = 210 cos56= 117.43 J

When the angle is 56° the work done is 117.43 J

You might be interested in

the first s-wave reaches a seismic station 22 minutes after an earthquake occurred. how long did it take the first p-wave to rea

Naddik [55]

The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

<h3>Time of travel of the P-wave</h3>

In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.

<h3>Relationship between speed and time</h3>

v ∝ 1/t

v₁t₁ = v₂t₂

t₁/t₂ = v₂/v₁

t₁/t₂ = 0.6v₁/v₁

t₁/t₂ = 0.6

t₁ = 0.6t₂

t₁ = 0.6 x 22 mins

t₁ = 13.2 mins

Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

Learn more about P-waves here: brainly.com/question/2552909

#SPJ1

7 0

2 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

--

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

--

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

--

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago

When the acceleration of a mass on a spring is zero, the velocity is at a

Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

$F=kx$

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

$F=ma$

where

m is the mass

a is the acceleration

So we can write the equation as

$ma=kx$ (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$ (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a maximum

When the velocity of a mass on a spring is zero, the acceleration is at a maximum

6 0

3 years ago

Dona rolls a marble with a speed of 5.0/m/s across a level table that is 1.6m above the floor. Upon reaching the edge of the tab

san4es73 [151]

Answer:

2.9 m

Explanation:

First find the time it takes to reach the floor.

y = y₀ + v₀ t + ½ at²

(0 m) = (1.6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.571 s

Next, find the distance it travels in that time.

x = x₀ + v₀ t + ½ at²

x = (0 m) + (5.0 m/s) (0.571 s) + ½ (0 m/s²) (0.571 s)²

x = 2.86 m

Rounded to two significant figures, the marble travels 2.9 meters in the x direction.

5 0

3 years ago

Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?

Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

$KE_1 = \dfrac{1}{2}mv^2$

$2000= \dfrac{1}{2}mv^2$ ........(1)

now, speed of the car is doubled

v' = 2 v

$KE_2 = \dfrac{1}{2}mv'^2$

$KE_2 = \dfrac{1}{2}m(2v)^2$

from equation (1)

$KE_2 = 4\times \dfrac{1}{2}m(v)^2$

$KE_2 = 4\times 2000$

$KE_2 = 8000\ J$

Hence, the Kinetic energy would be equal to 8000 J.

The correct answer is option A.

8 0

3 years ago