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MrMuchimi
3 years ago
5

A bullet is fired vertically upward a velocity of 80m/s to what height will the bullet rise above the point of projection​,note:

:this is a linear motion question
​
Physics
1 answer:
stellarik [79]3 years ago
3 0

Answer:

Height:

{ \bf{height =  \frac{ {velocity}^{2} }{gravity} }}

Formular:

{ \tt{height =  \frac{ {u}^{2} }{g} }}

Substitute:

{ \tt{height =  \frac{ {80}^{2} }{9.81} }} \\  \\ { \tt{height = 652.4 \: m}}

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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
MrMuchimi

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.

8 0
3 years ago
A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a
irinina [24]

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

Thus, the speed of the second satellite is less than the speed of the first satellite.

Learn more about speed.

brainly.com/question/7359669

#SPJ1

3 0
2 years ago
What is the
Len [333]

Answer:

d=5\ g/cm^3

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3

So, the density of the object is equal to 5\ g/cm^3.

6 0
3 years ago
If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement
alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

v(t)=3t^2-6t+24

In order to find the expression for the position of the car, we integrate this expression. We find:

x(t)=\int v(t) dt=t^3-3t^2+24t+C

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

x(0)=0^3-0+0+C=C

While the position after t = 3 s is

x(3)=3^3-3(3)^2+24(3)+C=72+C

Therefore, the displacement of the car in 3 seconds is

d=x(3)-x(0)=72+C-C=72

7 0
3 years ago
A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0
3 years ago
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