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mihalych1998 [28]
3 years ago
12

Playing catch with a friend in a moving train. When you toss the ball in the direction the train is moving, how does the speed o

f the ball appear to an observer standing at rest outside the train?
I understand basic relative motion, just the ball moving on another moving object confuses me!
Physics
2 answers:
likoan [24]3 years ago
5 0
Well it would simply appear faster because the speed of the ball and the speed of the train.
telo118 [61]3 years ago
5 0

Answer:

The speed of the ball appear faster when we observe it from ground frame at rest.

Explanation:

As we know that by relate motion concept

v_{AB} = v_A - v_B

here when ball is thrown in the direction of train motion with velocity "v" relative to the train then we will assume here that let say the train at that moment is running at speed v_T

now by above concept of relative velocity we will have

v = v_{ball} - v_T

now for the speed of ball

v_{ball} = v + v_T

so from this equation we can see that the actual speed of ball in ground frame will be more than the speed observed in train frame

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The water would be because however much salt you add the water rises
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The momentum of a 0.1 kg object traveling at 2000 m/s is 20,000 kg·m/s. True or False
Alina [70]

That's false.  

The definition of momentum is (mass) x (speed), so they must be multiplied.

"20,000 kg-m/s" has the correct units resulting from multiplication, but the number could only be the result of division.

3 0
3 years ago
You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex
sweet [91]
If the person if slowing down, then this would mean that this would actually be "negative acceleration". Sense this person first did a mile in 10 minutes, and then the next mile that this person did was actually 15 minutes, then this shows you that they actually went a lot slower. And this is why this would be a great example of "negative acceleration".
8 0
3 years ago
Read 2 more answers
What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?
liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)

where:

m_g = 3.0 kg is the mass of the gold bar

C_g=129 J/kg C is the specific heat of gold

T_g=99 C is the initial temperature of the gold bar

m_w = 0.22 kg is the mass of the water

C_w=4186 J/kg C is the specific heat of water

T_w=25 C is the initial temperature of the water

T_f is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C

5 0
3 years ago
A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d
Mnenie [13.5K]

Answer:

(C) 40m/s

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's  velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0
3 years ago
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