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Valentin [98]
2 years ago
10

Motion is changing all the time. When you stand up, begin to run, or turn a corner, your motion is changing. These changes in mo

tion all have something in common. They all start with a push or a pull. What happens when you push or pull an object? Make a claim about how a push or pull affects the motion of a water tube.
Physics
1 answer:
Andrej [43]2 years ago
6 0

Answer:

push on objects force is required. pull force is resli8

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Which statement does not describe the relationship between genes and
jek_recluse [69]

Answer: C. Inherited traits carry the instructions for individual genes.

Explanation: bejewels I know stuff. ☆ - ~ hope this helps

4 0
3 years ago
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The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m 828 m (2716.5 ft) and has more than 160 storie
Natasha_Volkova [10]

Answer:

h = height of the hotel room from the ground floor = 237.4m

Explanation:

Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh

PE1 is the potential energy of tourist at the ground floor

PE1 is the potential energy of tourist at the top (hotel room)

Given

PE1 = − 2.01 × 10⁵ J

PE2 = 0J

PE2 – PE1 = mgh

0 – (− 2.01 × 10⁵ J) = mgh

2.01 × 10⁵ J = 86.4×9.8×h

h = 2.01 × 10⁵/(86.4×9.8) = 237.4m

8 0
3 years ago
A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre
butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0
3 years ago
It takes 20 N of force to move a box a distance of 10 m. How much work is done on the box?
PIT_PIT [208]

Answer:

Ans is 200 J

Explanation:

Given:  Force = 20N

            Distance = 10m

Work done  = Force * displacement

                    =  20 * 10

                    =  200 J

3 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
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