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allsm [11]
3 years ago
14

Finish A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?

Physics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 6 \ meters}}

Explanation:

Work is the product of force and distance.

W=F*d

We know that 96 Joules of work were done and a 16 Newton force was applied to the object.

  • W= 96 J
  • F= 16 N

Substitute the values into the formula.

96 \  J= 16 \ N * d

First, let's convert the units. This will make cancelling units easier later in the problem. 1 Joule (J) is equal to 1 Newton meter (N*m), so the work of 96 Joules equals 96 Newton meters.

96 \ N*m= 16 \ N * d

Now, solve for distance by isolating the variable, d. It is being multiplied by 16 Newtons and the inverse of multiplication is division. Divide both sides of the equation by 16 N.

\frac {96 \ N*m}{16 \ N}= \frac{16 \ N *d}{16 \ N}

\frac {96 \ N*m}{16 \ N}=d

The units of Newtons cancel.

\frac {96}{16} \ m = d

6 \ m = d

The object moved a distance of <u>6 meters.</u>

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initial speed of the stuntman is given as

v = 28 m/s

angle of inclination is given as

\theta = 15 degree

now the components of the velocity is given as

v_x = 28 cos15 = 27.04 m/s

v_y = 28 sin15 = 7.25 m/s

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

\delta y = -25 m

\delta y = v_y * t + \frac{1}{2} at^2

-25 = 7.25 * t - \frac{1}{2}*9.8* t^2

by solving above equation we have

t = 3.12 s

Now in the above interval of time the horizontal distance moved by it is given by

d_x = v_x * t

d_x = 27.04 * 3.12 = 84.4 m

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.

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The correct option will be
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Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
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A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver
MissTica

Answer:

Approximately 2.31\; \rm m \cdot s^{-2} (assuming that the acceleration due to gravity isg = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Assuming that g = 9.81\; \rm m \cdot s^{-2} the weight on this 72-kg skydiver would be W = m \cdot g = 72 \; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 706.32\; \rm N (points downwards.)

Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.

Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:

\begin{aligned}F(\text{net force}) &= W - F(\text{air resistance})\\ &= 706.32\; \rm N - 540\; \rm N =166.32\; \rm N \end{aligned}.

Apply Newton's Second Law of motion to find the acceleration of this skydiver:

\begin{aligned}a &= \frac{F(\text{net force})}{m} \\ &= \frac{166.32\; \rm N}{72\; \rm kg} = 2.31\; \rm m \cdot s^{-2} \end{aligned}.

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