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1 year ago

A 25kg child resting at the top of a 2 meter slide has how much potential energy?

Physics

1 answer:

levacccp [35]1 year ago

3 0

Potential energy of a child at the top is 490 J.

Potential energy, U = mgh

U = potential energy

m = mass of an object, given = 25kg

g = acceleration due to gravity, given = 9.8 m/ $s^{2}$

h = height of an object, given = 2 meter

Put the values in potential energy formula, U = mgh

U = mgh

U = 25 × 9.8 × 2

U = 490 J

Hence, Potential energy of a child at the top is 490 J.

Potential energy is the type of energy, which is product of mass of an object, acceleration due to gravity and height of an object where it is placed. Its S.I. unit is joule and is represented by J.

Learn more about potential energy here:- brainly.com/question/14427111

#SPJ1

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What class lever is a

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should be 1

Explanation:

because what do you use in class and has a lever

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3 years ago

A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc

Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity $v_1$ in time $t_1$

$\frac{d}{2}=v_1t_1----1$

from B to C

distance traveled is 0.5 d with $v_2$ and $v_3$ velocity for half-half time

$\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2$

divide 1 and 2 we get

$\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}$

$\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}$

Now average velocity is given by

$v_{avg}=\frac{d}{t_1+t_2}$

taking $t_1$ common

$v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}$

$v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}$

$v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}$

6 0

3 years ago

Guys I really need help with these 2 questions , it's for my final plz help asap

mars1129 [50]

Answer:

(1) Initial speed, $u=0$

Final speed, $v=165.76m/s$

Average speed, $v_a_v_g=82.87m/s$

(2) Force of gravity, $F_g=12.8\times10^1^5N$

Explanation:

(1)

Given,

Distance, $S=300meter$

Time, $t=3.62second$

It is given that drag racer started at rest.

So Initial speed, $u=0$

Using Newton's second equation of motion,

$S=ut+\frac{1}{2}at^2\\300=0+\frac{a\times3.62^2}{2} \\a=45.79m/s^2$

Newton's first equation of motion,

$v=u+at\\=0+45.79\times3.62\\=165.76 m/s$

So, Final speed, $v=165.76m/s$

Average speed is defined as totle distance divided by totle time.

$v_a_v_g=\frac{S}{t}\\=\frac{300}{3.62} \\=82.87m/s$

So, Average speed, $v_a_v_g=82.87m/s$

(2)

Gravitation: It is the natural phenomenon in which two different bodies attract each other by virtue of their masses.

According to Newton's law of gravitation, the force of attraction between two bodies is directly proportional to the masses of the bodies and inversely proportional to square of distance between centers of mass of the bodies.

$F_g\propto\frac{m_1m_2}{r^2} \\F_g=G\frac{m_1m_2}{r^2}$ where $G$ is constant of proportionality and known as gravitation constant.

Given,

Mass of Jupiter, $m_1=1.9\times10^2^7kg$

Mass of Ganymede, $m_2=1.48\times10^2^3kg$

Distance between their centers of mass, $r=1.21\times10^1^2meter$

$F_g=G\frac{m_1m_2}{r^2}\\=\frac{6.67\times10^-^1^1\times1.9\times10^2^7\times1.48\times10^2^3}{(1.21\times10^1^2)^2} \\=12.8\times10^1^5N$

So, Force of gravity, $F_g=12.8\times10^1^5N$

7 0

3 years ago

3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a

34kurt

Answer:

The friction force and the x component for the weight should be the reaction forces that are opposite and equal to the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.

Explanation:

When the locomotive starts to pull the train up, appears two reaction forces opposed to the action force in the direction of the move.

The first one is due to the friction between the wheels and the ground, it will be the friction force (Fr):

Fr = μ*Pₓ =μmg*sin(φ)

where μ: friction dynamic coefficient, Pₓ: is the weight component in the x-axis, m: total mass = train's mass + locomotive's mass, g: gravity, and sin(φ): is the angle respect to the x-axis.

And the second one is the x component for the weight (Wₓ):

Wₓ = mg*cos(φ)

where cos(φ): is the angle respect to the y-axis.

These two forces should be the same as the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.

3 0

3 years ago

A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of soun

ella [17]

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

4 0

3 years ago