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borishaifa [10]
3 years ago
13

Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

y of 24 m/s?
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

So, the final velocity is

v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

\Rightarrow v=-15.24 m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

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The dimensions of a block that weighs 30 N are 0.23 mx 0.12 mx 0.075 m.
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Answer:25000/23 N/M^2

Explanation:

P=f/A=30/(0.23×0.12)=25000/23

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A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an a
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RESULT: Twalk= 3.64 hr

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2 years ago
I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit
Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

M=5.97\cdot 10^{24} kg (mass of the Earth)

m=3.0 kg (mass of the box)

r=R=6.37\cdot 10^6 m (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N

2) g=9.8 m/s^2

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol g.

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

F=ma

where a is the acceleration of the object. Re-arranging,

a=\frac{F}{m}

which is exactly equal to the quantity we have calculated above.

5 0
3 years ago
a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity
stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

V_{ox} = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = V_{ox}   t

80 = (v Cos60) (t)

t = 160/v                                         eq-1


consider the motion in vertical direction :

Y = vertical displacement = 20 m

V_{oy}  = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = V_{oy}  t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0
3 years ago
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