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Tasya [4]
3 years ago
15

Hey guys please help ✋✋

Physics
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

by using mearsung scale

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Objects that rest have no forces upon them
Pani-rosa [81]

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

6 0
3 years ago
1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car e
tankabanditka [31]

Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

Acceleration = a = ?

F = ma

a = F/m

a = 2000/ 1000

a = 2 m/s^{2}

Q2: Solution:

Change in velocity = Δv = ?

acceleration = change in velocity / time

a = Δv/t

Δv = axt

Δv = 2 x 10

Δv = 20 m/s

Q3: Solution:

Impulse = I = ?

Impulse = Force x time

I = 2000 x 10

I = 20000 Ns

Q4: Solution:

Change in Momentum = Δp = ?

Δp = mΔv

Δp = 1000 x 20

Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

Δp = m x (vf-vi)

Δp = 1000 x (vf - 0 )

20000 = 1000 x vf

vf = 20000/1000

vf = 20 m/s

Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

Δp = 1000 x (0-20)

Δp = -20000 kgm/s

Q9: Solution:

Impulse = Δp

Impulse = -20000 Ns

Q10: Solution:

Impulse = ?

Impulse = F x t

F = impulse/t

F = -20000/4s

F = -5000 N

Q11: Solution:

F = ma

a = ?

a = F/m

a = -5000/1000

a = -5m/s^{2}

6 0
3 years ago
How and why does the distance between 2 electrodes affect the rate of electrolysis? ...?
Rasek [7]
 <span>Nothing, in terms of the chemistry. 

The distance between the electrodes affects the electrical resistance very slightly. Increasing the distance increases the resistance and reduces the current slightly, which reduces slightly the amount of product. 

For most practical applications, for electrolysis done in a beaker, varying the distance between the electrodes will make little difference. 

Increasing the concentration of the electrolyte will increase the current flow because there are more charged particles to carry charge, and increase the product yield.</span>
4 0
3 years ago
Tell me all about acids and bases (NO GOOGLE SEARCH)
aleksklad [387]
Acids are danger so stay away
7 0
4 years ago
Read 2 more answers
If two ends A and B are connected through a metallic wire. If electrons flows from point B to point A. What can you say about th
Ivan

Answer:

Point A is at higher potential than point B

Explanation:

Electrons are negatively charged - this means that they are attracted by positive charges and repelled by negative charges.

This also means that they tend to move in a direction opposite to the electric field lines (because electric field lines point away from a positive charge and toward a negative charge). So, they also tend to move from a point at lower potential to a point at higher potential.

In this problem, the electrons flow from point B to point A: therefore, point A must have higher potential than point B.

6 0
3 years ago
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