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Gnoma [55]
3 years ago
12

what do you understand by gravitational potential energy? what is the increase in the gravitational potential energy of a 20 kg

block which is lifted 10m from the ground?​
Physics
2 answers:
enyata [817]3 years ago
5 0

Answer:

  • energy possessed by any object due to its position in gravitational field
  • it is given by the equation P.E.= mgh where m = mass ,g= gravitational force of earth ( 9.8m/s^2) and h= hight of the object
  • there for for above problem m = 20kg , h= 10 m and g= (9.8m/s^2)
  • p.E. = 20×10×9.8 =1960 jules

Explanation:

hope it helped u buddy :-)

Rainbow [258]3 years ago
5 0

Answer:

grams

Explanation:

mass

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A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of
inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

v^2-u^2=2ad

-u^2=2ad

a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2

The frictional force on the block is given by the formula as :

F = ma

F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

8 0
4 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
A conveyor belt at a recycling plant launches bottles and bottle caps into the air, so that an automatic image recognition devic
dem82 [27]

(a) The initial speed at which the bottles are launched is 4.27 m/s.

(b) The horizontal displacement at which the bottle land is 1.75 m.

<h3>Initial speed of the bottle</h3>

The initial speed of the bottle is calculated as follows;

T = \frac{2usin\theta}{g}

where;

  • T is time of flight
  • u is the initial speed

2usinθ = Tg

u = Tg/(2sinθ)

u = (0.5 x 9.8)/(2 x sin35)

u = 4.27 m/s

<h3>Horizontal displacement of the bottle</h3>

X = u²sin(2θ)/g

X = (4.27² x sin(70))/(9.8)

X = 1.75 m

Learn more about projectile here: brainly.com/question/12870645

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The driving force for an electric current is called
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Its called electro motive force . let me know if its right

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Talja [164]

Answer:

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Explanation:

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