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Zinaida [17]
3 years ago
8

A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th

e ground? (Disregard air resistance. a = -g = -9.81 m/s.)
Physics
1 answer:
Bad White [126]3 years ago
8 0

Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground  = 5.7s

Unknown:

Final velocity of the coin  = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

             V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

               V  = 0 + 9.81 x 5.7

               V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

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A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

5 0
2 years ago
Two converging lenses are separated by 27.0 cm. The focal length of each lens is 8.90 cm. An object is placed 33.0 cm to the lef
Ymorist [56]

Answer:

1/i + 1/o = 1/f     thin lens equation

i = 33 * 8.9 / (33 - 8.9) = 12.2 cm  to right of first lens

27 - 12.2 = 14.8 cm to left of second lens

i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens

7 0
2 years ago
Name the following ionic compounds: BeCl2​
tatyana61 [14]

Answer:

Beryllium chloride

Explanation:

I hope this helps!

Can I get branliest?!

4 0
3 years ago
A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l
aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass =  5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate  

and

here work need to lift cable is

w = (10Δy )(9.8 y )  j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= \int\limits^{10}_0 {98y} \, dy    

= [49 y^2]^{50}_0

= 2450J

so lifting 5000 lb wrcking 50 m  required additional 5000 + 2450

so total work is = 52450 J

3 0
3 years ago
1. How much power does a light bulb contain if it does 600J of work in 5 seconds?
dimaraw [331]

Answer:

120 watts

Explanation:

#1: 120 watts

#2: 667 watts

#3: 3 watts

#4: i forgot how to do this one

5 0
2 years ago
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