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4 years ago

HELP ASAP! GIVING BRAINLIEST

Physics

2 answers:

Sergio [31]4 years ago

8 0

Answer:

Sound energy..

Explanation:

alina1380 [7]4 years ago

5 0

Answer: Kinetic Energy?

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For the velocity-time graph showns, which statement describes what happens to the velocity between approximately 11 and 15 secon

dalvyx [7]

Answer:

C) The lander’s velocity decrease toward the reference.

Explanation:

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3 years ago

Why is physics the hardest subject

horrorfan [7]

Physics can demand that you start with a specific result and make general rules. Not reading the text and not solving exercises makes understanding almost impossible. It also involves a lot of math which most people do not like.

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A student claims that since the three phases have different amounts of energy, molecules in substances gain or lose energy only

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if heat transferred to any phase that phase has had a heat transfer

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3 years ago

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A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after

Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g= $9.81ms^{-2}$

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

$v=u+at$

Using the formula and substitute the values

$0=24.1-9.81t$

Where g is negative because motion of ball is against gravity

$24.1=9.81t$

$t=\frac{24.1}{9.81}=2.46s$

3. $y=y_0+ut+\frac{1}{2}at^2$

Using the formula

$y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2$

$y=30.1 m$

4.Time of flight for red ball=3.77-2.9=0.87s

$y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2$

$y'=22.0m$

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

$0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2$

$0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2$

$0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2$

$0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t$

$-2.86105=-2.851t$

$t=\frac{2.86105}{2.851}=1.004 s$

Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.

8 0

3 years ago

If you start rolling down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill ypu KE

Serjik [45]

Answer:

$v=\sqrt{(2x/m)}\ \ m/s$

Explanation:

Potential Energy= Kinetic Energy

Let $x$ be the value of Kinetic Energy.

We know that

$PE=KE=x\\x=\frac{1}{2}mv^2\\$

Make $v$ the subject of the formula to get speed at the bottom of the hill.

$v^2=2x/m\\v=\sqrt{2x/m}$

8 0

3 years ago