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den301095 [7]
2 years ago
13

A soccer ball is dropped from the height of 2m and falls to the floor. What is the relationship of the potential and kinetic ene

rgies of the ball as the ball falls to the floor?
Physics
1 answer:
Alexandra [31]2 years ago
4 0

We know , gravitational force is a conservative force .

Therefore , total energy of the system remains constant .

P.E_i+K.E_i=P.E_f+K.E_f

Let , mass of object is m.

So ,

mg(2)+0=0+\dfrac{mv_f^2}{2}\\\\v_f=\sqrt{4g}\\\\v_f=\sqrt{4\times 9.8}\ m/s\\\\v_f=6.26\ m/s

Therefore , the velocity at floor is 6.26 m/s .

Hence , this is the required solution .

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A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
2 years ago
PLEASE HURRY!
zloy xaker [14]

PRETTY SURE ITS A AND C BUT I LOOKED IT UP AND IM ALMOST POSITIVE THARS IT

8 0
3 years ago
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PLEASE HELP ME WITH A PHYSICS QUESTION!!!!!!!!!!
stira [4]

Answer:

C. 3.00 s

Explanation:

Given:

Δy = 1.80 m − 46.0 m = -44.2 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 3.00 s

6 0
3 years ago
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An eagle flies from its perch in a tree to the ground to capture and eat its prey. describe its energy transfromation
ddd [48]
10% energy is transferred from the prey to eagle and 90% energy is lost in the enviorment .
7 0
3 years ago
Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/
Pani-rosa [81]

Answer:

equation of  motion for Bill is

y(t) = 4.9t^2

equation of  motion for Ted is

y(t) = 2 + (-4.2)(t) + 4.9t^2

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of  motion for Bill is

y(t) = y_0 +v_0 t +\frac{1}{2}gt^2

y(t) = 0 + 0(t) +\frac{1}{2}gt^2

y(t) = \frac{1}{2}\times (9.8t)^2

y(t) = 4.9t^2

equation of  motion for Ted is

y_0 = 2m -1m = 2m

y_0 = -4.2 m/s

y(t) = y_0 +v_0 t +\frac{1}{2}gt^2

y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2

y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2

y(t) = 2 + (-4.2)(t) + 4.9t^2

8 0
3 years ago
Read 2 more answers
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