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Dimas [21]
2 years ago
6

11. An object moves in circular path with constant speed

Physics
1 answer:
Temka [501]2 years ago
6 0

Answer:

B. Is its acceleration constant

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by v_{max}=A\omega where A is amplitude and \omega is angular velocity

Angular velocity is given by \omega=\sqrt{\frac{k}{m}}  where k is spring constant and m is mass

So v_{max}=A\sqrt{\frac{k}{m}}

A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m

3 0
3 years ago
Read 2 more answers
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)
strojnjashka [21]

Answer: not sure

Explanation:

4 0
2 years ago
Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0
2 years ago
I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow
Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0
3 years ago
Can someone help me with this please?
Slav-nsk [51]

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

8 0
3 years ago
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