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Hunter-Best [27]
3 years ago
13

Which describes the relationship between potential and kinetic energy of a ball thrown up in the air as it falls back to the gro

und?
A.
Gravitational potential energy increases; kinetic energy increases.


B.
Kinetic energy equals gravitational potential energy.


C.
Kinetic energy increases; gravitational potential energy decreases.


D.
The mass of the ball is required to answer this.
Physics
1 answer:
Gekata [30.6K]3 years ago
7 0
The answer is c, because ball is falling so its gravitationl potential energy decreases, but it kinetic energy increases. Energy is always conserved.
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In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector
icang [17]

Answer:

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx  = cx and ay + by = cy. From this two equations we can obtain B.

bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m

by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m

Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.

Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) =  26.58 m.

So the magnitude of B is 26.58 m.

7 0
3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

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3 years ago
If a block is in equilibrium the magnitude of the block's acceleration is
kaheart [24]

zero.

from newton's first law of motion

3 0
3 years ago
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A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in
Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

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3 years ago
List what sources of uncertainty go into calculating the wavelength of the laser (no explanation necessary here). (b) Accurately
laiz [17]

Answer:

thanks for da 5points hoi

Explanation: thanks dawg

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3 years ago
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