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3 years ago

If a dog ran at 5 m/s how far would it run in 45 s

Physics

2 answers:

saveliy_v [14]3 years ago

8 0

Answer:

225 meters

Explanation:

If it is running 5 meters per seconds, and it ran 45 seconds, you would need to multiple the seconds it ran by the speed per second: 5 • 45

5 • 45 = 225

Virty [35]3 years ago

3 0

Answer:

225 m//s

Explanation:

It ran 5m/s so 5x45=225

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If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13

RUDIKE [14]

According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell.

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
= 20 / (0.15 · 9.8)
= 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0

3 years ago

3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl

Marina86 [1]

The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.

6 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago

Your car is traveling at 70/mi/h (1000ft/s) your sneeze and your eyes close for 0.33 seconds how far did you travel during your

Novay_Z [31]

If you're moving at 70 mi/hr, then you cover 33.88 feet in 0.33 sec. If you're moving at 1,000 ft/sec, then you cover 33 feet in 0.33 sec. 70 mph and 1,000 fps are not equivalent. 1,000 fps is about 682 mph, whereas 70 mph is about 103 fps.

6 0

4 years ago

The moon has a smaller mass than the Earth. If

ArbitrLikvidat [17]

Answer:

Decrease

Explanation:

If you were on the Moon, which has significantly less mass than the Earth, your weight would: decrease

The point that seemed to be giving me a complicated time was being able to distinguish the difference and meaning of weight and mass and being able to apply that to a problem. I kept mixing up the definitions. For example in homework 3.1, one question asked:

If you were on the moon, which has significantly less mass than the earth, your mass would:

a. increase

b. decrease

c. stay the same

d. become zero

The definition of mass is the amount of matter in an object. The definition of weight is the amount or unit of force. For me, I just had to remember that when it asked about weight, it wasn’t referring how heavy an object is. After I was able to recognize that when it came to weight, questions became easier.

The final and correct answer was decreasing. The answer is because the Moon’s mass is less. This means the gravitational force is less on your body, therefore, your mass is going to be lighter

5 0

3 years ago