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3 years ago

11

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding. Disregarding friction, what is the velo

city of the Mr. Miles at the bottom of the slide? (g = 9.81 m/s2.)

1 answer:

lilavasa [31]3 years ago

5 0

Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

$v=\sqrt{2gh}$

g is the acceleration due to gravity

$v=\sqrt{2\times 9.8\times 15}$

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

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Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force

Learn more about gravitational force and electric force:

brainly.com/question/1724648

brainly.com/question/12785992

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

5. What happens to the arrangement of water molecules as ice melts?

m_a_m_a [10]

Answer: I am pretty sure the answer is B

Explanation: If not sorry bro.

7 0

3 years ago

In a given chemical reaction, the energy of the products is less than the energy of the reactants. Which statements is true for

Inessa05 [86]

Answer:It is an Endothermic reaction

Explanation:

Exothermic reaction

An exothermic reaction is one that releases energy in the form of heat or light.

Endothermic reaction

An endothermic reaction is one that absorbs energy in the form of heat or light.

7 0

3 years ago

Read 2 more answers

A submarine can detect an approaching enemy submarine using sonar. If a stationary submerged submarine emits a 100kHz tone and r

kotegsom [21]

Answer:

B) 1.5 m/s

Explanation:

The apparent frequency will be enhanced due to Doppler effect

If f be the apparent frequency , F be the real frequency , V be the velocity of sound and v be the velocity of approaching submarine then f is given by

f = F \frac{V+v}{V-v}\\

\frac{f}{F} =\frac{V+v}{V-v}\\

\frac{f}{F}-1 =\frac{V+v}{V-v}-1\\

\Delta f = \frac{2vf}{V-v}\\

200=\frac{2\times v\times 100\times 1000}{1482-v}\\

v=1.48 m/s

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3 years ago