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3 years ago

What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?

Physics

1 answer:

Gekata [30.6K]3 years ago

7 0

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

$p = mv$

Let's replace what we know and solve for whatever's left

$100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s$

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If the velocity of the particle is uniform, then the path of the particle must be a

lianna [129]

Answer:

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3 years ago

Can instantaneous velocity ever be negative?

Nana76 [90]

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3 years ago

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2× $10^{-6}$ J

(b) KE = 2× $10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0

3 years ago

Divers in Acapulco dive from a cliff that is 65 m high. If the rocks

Alex777 [14]

Answer:

Cool question! First step is to find the time taken to fall

, then to find the horizontal velocity needed to cover

in that time. In this case the answer is

7.0

−

Explanation:

This is a less typical projectile motion question, but it's still projectile motion. This means the horizontal and vertical directions can be considered separately. We assume that the initial vertical velocity,

−

, and we are trying to find the required initial horizontal velocity,

To find the time taken to fall

Since

, we can rearrange this to:

√

⋅

9.8

3.41

The horizontal velocity will be constant (ignoring air resistance), so to cover

3.41

will be given by:

→

3.41

7.0

−

Answer link

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2 years ago

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My name is Ann [436]

Um he made "determine the state of matter" all caps?

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3 years ago