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3 years ago

A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?

Physics

2 answers:

marusya05 [52]3 years ago

7 0

KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s

anyanavicka [17]3 years ago

5 0

Answer:

Kinetic energy = 225 J

Explanation:

As we know that kinetic energy is given as

$KE = \frac{1}{2}mv^2$

now here we know that

m = 2 kg

v = 15 m/s

now from above formula we know

$KE = \frac{1}{2}(2)(15)^2$

now we have

$KE = 225 J$

so kinetic energy of the moving ball is 225 J

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3 years ago

What is the change in velocity of a 22-kg object that experiences a force of 15 N for

vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

${ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}$

From first Newton's equation of motion:

${ \bf{v = u + at}}$

Change = v - u:

${ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}$

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3 years ago

A car driving at 20 m/s accelerates continuously 2m/s2 for 3 seconds. What is its final velocity

Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0

2 years ago

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

3 0

3 years ago

A bottle rocket takes off with a = 34.5 m/s2. It

sergiy2304 [10]

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

Take them square root of both sides

t=√(11.59)

t=3.4 seconds

8 0

3 years ago