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Marrrta [24]
3 years ago
14

Max (15 kg) and Maya (12 kg) are ice-skating on a frozen pond. While standing at the center of the pond, Maya throws a 1.5-kg sn

owball at Max and, as a result, recoils away from Max at 2.5 m/s. With what speed did Maya throw the snowball at Max
Physics
1 answer:
pychu [463]3 years ago
4 0

Answer: The speed at which Maya threw the snowball at Max is 20m/s

Explanation:

Given that

​Mass of snowball=   1.5-kg.

Mass of Maya= 12kg

Initial  speed of Maya=2.5m/s

We use the law of conservation of  momentum to calculate the  speed of the snowball which depends on its own mass,  the mass of Maya who threw the snowball and the initial velocity at which the snowball was thrown by Maya

Ms x Us = Mm x Um

Mass of snowball  x Initial speed of Snow ball =Mass of Maya x Initial  speed of Maya

1.5kg  x Initial speed of Snow ball =12 kg x 2.5m/s

Initial speed of Snow ball=( 12 kg x 2.5m/s) /1.5kg

​ Initial speed of Snow ball  =20m/s

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Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest. Which best describes h
meriva

Answer:

8.1km/h Northwest

Explanation:

The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.

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For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?
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4 0
3 years ago
Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

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Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

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4 0
3 years ago
the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

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F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



6 0
3 years ago
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