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natima [27]
2 years ago
6

Urgent i need help with these questions

Physics
2 answers:
Anton [14]2 years ago
8 0

Answer:

Question 1: 25

Explanation:

i cant do the rest soz

Harrizon [31]2 years ago
4 0

Answer:

1. 25%

2. You can move the skater higher up to increase the potential energy.

3. The kinetic energy increases because there is more motion.

Explanation:

1. If the total kinetic and potential energy is 100%, and we know that the kinetic energy is 75%, then we can find the potential by finding the missing value of this equation. Potential energy + kinetic energy =total energy.

? + 75%=100%

2. The higher an object is, the more potential energy it has because of the increase of its gravitational energy.

3. When an object is higher, there is more potential energy. As the skater goes down, the potential energy decreases which in turn increases the kinetic energy.

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3 years ago
A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered
mart [117]

Answer:

\omega_f=12.2954 \,rad.s^{-1}         i.e.

Frequency=1.9569\,rev.s^{-1}

Explanation:

Given:

angular speed,\omega_i =2\times 2\pi \,rad.s^{-1}

mass of the disk, M=4.9\,kg

radius of the disk, R=0.2\,m

mass of the chunk, m=2.7\,kg

radius of the chunk, r=0.04\,m

We know that the angular momentum is given by:

L=I.\omega

and moment of inertia for a disc:

I=\frac{1}{2} m.r^2

According to the conservation of angular momentum, the final angular momentum is equal to the initial angular momentum.

\frac{1}{2} \times M.R^2\times \omega= \frac{1}{2} \times (M.R^2+m.r^2) \omega_f

4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f

\omega_f=12.2954 \,rad.s^{-1}

Frequency=1.9569\,rev.s^{-1}

4 0
4 years ago
How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?
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Answer:

C. The block and the sphere would have the same weight.

Explanation:

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Where the displacement is a maximum the pressure is a minimum.
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