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lana66690 [7]
3 years ago
9

A 4 kg block is pushed up an incline that makes a 30° angle with the horizontal, as shown in the figure. Once the block is pushe

d a distance of d = 5.0 m up the incline, the block remains at rest.
What is the approximate change in the gravitational potential energy of the block-Earth system when the block is held at rest compared to its original location at the bottom of the incline?
A) OJ
B) 100 J
C) 100/33 J
D) 200 J
Physics
1 answer:
Sophie [7]3 years ago
8 0

Answer:

  B)  100 J

Explanation:

Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...

  ∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J

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The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
AnnyKZ [126]

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

5 0
3 years ago
(b) Two containers made of insulating material contain the same volume of water at room
masha68 [24]

Answer:

the volume of liquid decreased due to evaporation from the exposed free surface of water so molecules got evaporated .

evaporation occurs at room temperature.

5 0
2 years ago
A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl
sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

\theta=18.5^{\circ}

Tension=140 N

Initial velocity of wagon=u=0

Displacement=s=80 m

Net force applied  on wagon=F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N

By using g=9.8m/s^2

a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2

We know that

v^2-u^2=2as

Using the formula

v^2=2\times 0.39\times 80

v=\sqrt{2\times 0.39\times 80}=7.9m/s

5 0
3 years ago
Trace fossils would include all of the following EXCEPT
Vadim26 [7]

Answer:

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Explanation:

6 0
3 years ago
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

5 0
3 years ago
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