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ss7ja [257]
3 years ago
9

Trolley A has a mass of 2kg and is moving at 2m/s. It collides with trolley B, which has the same mass but is stationary. What i

s the momentum of the two trolleys after the collision?
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
Assuming your two trolleys stick together after the collision, your momentum equation would be M1V1+M2V2=(M1+M2)V3 and since V2 is 0, then 2x2=(2+2)V3, thereore the final velocity would be 1, wich would make the final momentum 4
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What is the SI unit for moment​
natali 33 [55]

Answer:

Newton Metre

Explanation:

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3 years ago
"If you double the wavelength of a wave on a particular string", what happens to the wave speed v and the frequency f ? (i) v do
nadezda [96]

Answer:

v doubles and f is unchanged

Explanation:

According to the formula v = f¶

Where v is the velocity of the wave

f is the frequency

¶ is the wavelength

Velocity is directly proportional to wavelength. Direct proportionality shows that increase in velocity will cause an increase in the wavelength and decrease in velocity will also cause a decrease in wavelength with the frequency not changing since the velocity and wavelength are both increasing and decreasing at the same rate.

According to the question, if the wavelength is doubled, the velocity (v) will also double while the frequency (f) remains unchanged.

5 0
3 years ago
Read 2 more answers
Help please!
JulsSmile [24]

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

7 0
2 years ago
I need help plz!!! Please help...
RSB [31]

Answer: 60

Explanation:

4 0
2 years ago
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (F_{t}), vertical gravitational force (F_{g}) and vertical normal force (F_{n}), due to the surface. Since there is no vertical movement, F_{g} and F_{n} cancels it out. So, for this block, net force is horizontal due to the rope F_{t}.

The block of mass m is hanging from the pulley, so there is the force of the rope (F_{t}) and the gravitational force (F_{g}). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

F_{r}=m.a

a=\frac{F_{r}}{m}

a=\frac{18.8}{3.6}

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

F_{t}-m.g=m.a

18.8-9.8m=5.22m

15.02m=18.8

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: F_{g}=m.g.

The weight for the hanging block is

F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0
3 years ago
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