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O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to

Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

3 0

3 years ago

What do we call magnets that are only magnetized when they are within an existing magnetic field?

Zigmanuir [339]

Answer: I think the answer is D

Explanation: N/A

3 0

3 years ago

Anyone know what type of circuit this is?

Oksana_A [137]

Answer:

C

Explanation:

BECAUSE ITS GOING ON AND ON IF ITS NOT CORRECT I WILL VOTE YOU BRAINLEST ON MY QUESTION

3 0

3 years ago

HELP. A river flows South with a speed of 8m/s. From the shore, you see a boat moving South with a speed of 10 m/s. From their f

riadik2000 [5.3K]

Answer:

18 m/s

Explanation:

Given that,

Speed of river is 8 m/s due South

From the shore, you see a boat moving South with a speed of 10 m/s. Both boat and river are moving in same direction. The resultant velocity will add up.

v = 10 m/s + 8 m/s

v = 18 m/s

Hence, the rowers are moving with a speed of 18 m/s.

3 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

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