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mrs_skeptik [129]
3 years ago
15

Question 3 (1 point)

Physics
1 answer:
ASHA 777 [7]3 years ago
7 0

Answer:

twice as much work

Explanation:

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A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what
belka [17]

Answer: 2 seconds

Explanation:

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3 years ago
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Which of these cultures did NOT make major contributions to astronomy?
Svetlanka [38]
I can't particularly place what Icelandic contributed.

But every  other culture contributed majorly.

So I would go for option b.
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3 years ago
This version of Einstein’s equation is often used directly to find what value?
Setler [38]
This version of Einstein’s equation is often used directly to find what value? E = ∆mc2

Answer: This version of Einstein’s equation is often used directly to find the mass that is lost in a fusion reaction. Therefore the correct answer to this question is answer choice C).

I hope it helps, Regards.
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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th
ehidna [41]

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

5 0
3 years ago
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