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kobusy [5.1K]
3 years ago
14

Plz help w answer 1:/ confused ash

Physics
2 answers:
amm18123 years ago
8 0
Yeah that’s the same question I was asking

Galina-37 [17]3 years ago
5 0

Answer:

I would say d I had the same question yesterday and I got it correct so hope that helps

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What is energy and types of energy<br>​
pishuonlain [190]
Energy- the ability to do work/how things can change and move

Types
Potential Energy
Kinetic Energy
Nuclear Energy
Mechanical Energy
Sound Energy
Heat
8 0
3 years ago
Read 2 more answers
What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25
Rom4ik [11]

Acceleration of the ball is 50 m/s^2

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
stepladder [879]
The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
3 0
3 years ago
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
3 years ago
Failure to accomplish Erikson’s psychosocial task of late adulthood leads to despair.
Y_Kistochka [10]
False is the answer
6 0
3 years ago
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