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klemol [59]
3 years ago
10

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

Physics
1 answer:
madam [21]3 years ago
3 0

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

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Mady has an infinite number of balls and empty boxes available to her. The empty boxes, each capable of holding four balls, are
True [87]

Answer:

The number of balls in the box at 2010th step is 6 balls

Explanation:

Since the boxes are arranged from left to right, and since the empty boxes are capable of holding 4 balls at once.

Therefore, her steps would follow this pattern ;

00001 - 1st Step

0002 - 2nd Step

0003 - 3rd Step

0004 -4th Step

0010 - 5th Step

0011 - 6th Step

0021 - 7th step

0031 - 8th step

0041 - 9th step

Looking at this pattern, it is obvious this digits are in base 5 because once we count to 4 in a unit, we move to the next at 1.

Thus,at Mady's 2010th step, we'll convert to base 5,thus 2010 in base 5 is calculated as;

2010 ÷ 5 = 402 r 0, so Base 5 is now: _ _ _ _ 0.

402 ÷ 5 = 80 r 2, so Base 5 is now: _ _ _ 2 0.

80 ÷ 5 = 16 r 0, so Base 5 is now:

_ _ 0 2 0.

16 ÷ 5 = 3 r 1, so Base 5 is now: _ 1 0 2 0.

3 ÷ 5 = 0 r 3, so Base 5 is now: 3 1 0 2 0.

So 2010 in base 10 is 31020 in base 5. Thus,the number of balls in the box at 2010th step is equal to 3 + 1 + 0 + 2 + 0 = 6balls

3 0
3 years ago
A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around
kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

h = 2.86 m

8 0
3 years ago
Read 2 more answers
Can some please help me with this?
Lerok [7]

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I would say that these would be your correct answers, btw I'm doing something that is close to the same right now

Hope this helps :)

6 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
As a pelican flies through the air, it flaps its wings, thereby pushing down on the air below. What is the reaction force?
madam [21]

Answer:

the reaction force in this situation would be B

Explanation:

The action is the wings pushing down whilst the reaction is the air pushing up which allow the bird to fly .

plz mark brainliest to help me lvl up :P

7 0
3 years ago
Read 2 more answers
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