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MAVERICK [17]
3 years ago
10

What is studying or working i only one field (this is for science 7th grade)

Physics
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

Studying is reviewing based on what you learned .Working is doing what you got to do while learning.

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What is the impulse of a dodgeball hitting a person in the face if it's initial velocity was 10m/s and it's velocity after hitti
Alborosie

Answer:)

Explanation:)

6 0
2 years ago
Can a body have zero velocity and finite acceleration?Explain​
sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0
3 years ago
A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0
3 years ago
A person is pulling a freight cart with a force of 58 pounds. how much work is done in moving the cart 70 feet if the cart's han
Kobotan [32]

<span>The person is dragging with a force of 58 lbs at an angle of 27 degrees relating to the ground. We want to use cosine function to look for the horizontal force component. And then we can compute for W = (Horizontal Force) x (Distance). We want the horizontal force component since that is the component that is parallel to the direction the cart is moving. </span><span>

(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal force component.) 
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>

6 0
3 years ago
6. A 50 N block is raised 2 m. If the net work done on the block is 50 J, what is the applied force on the block?
geniusboy [140]

Answer:

F = 75[J]

Explanation:

We know that work is defined as the product of force by distance.

In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

W = -(F_{weight*d})+(F_{upward}*d)

where:

W = work = 50 [J]

d = distance = 2 [m]

Fweight = 50 [N]

Fupward [N]

Now replacing:

50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]

5 0
3 years ago
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