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3 years ago

HELP MEEE

Physics

1 answer:

daser333 [38]3 years ago

6 0

Answer:

The answer should be True, but it's False on Edge 2021

Explanation:

이것이 도움이 되었기를 바랍니다.

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If a planet's orbital speed is 20 km/s when it's at its average distance from the sun which is most likely orbital speed when it

Alona [7]

Answer:25km/s

Explanation:

8 0

3 years ago

Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

Snowcat [4.5K]

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

5 0

3 years ago

Read 2 more answers

The direction of an electric field is the direction (5 points)

vlabodo [156]

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.

5 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

A 1.90-kg mass vibrating up and down on the end of a vertical spring has a maximum speed of 2.30 m/s. What is the total potentia

Pepsi [2]

Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg

Therefore the maximum kinetic energy of motion is given by

Kinetic Energy, KE = $\frac{1}{2} mv^{2}$

Where,

m = Attached vibrating mass = 1.90 kg

v = velocity of the string = 2.3 m/s

Therefore Kinetic Energy, KE = $\frac{1}{2}$ ×1.9×2.3² = 5.0255 J

From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion

Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion

That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

Total PE = Maximum KE = 5.0255 J.

6 0

3 years ago