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3 years ago

A 600kg car is at rest , and then accelerates to 5n/s what is the original kinetic energy?

1 answer:

7 0

M = 600 kg
u = 0 (as the car was at rest initially)
v = 5 m/s
Initial kinetic energy = ½mu²
= ½×600×0
= 0
Final kinetic energy = ½mv²
= ½×600×25
= 7500 J

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A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is th

tatiyna

Answer:

F = - K x

a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2

b) ω = (K/m)^1/2 angular frequency of SHM

ω = (133 / 1.3)^1/2 = 10.1 / sec

f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec

P = 1/f = .0157 sec

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3 years ago

The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the gr

WINSTONCH [101]

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

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3 years ago

A student visits the beach and wants to explore how landforms, such as sand dunes, are the result of changes caused by wind. wha

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3 years ago

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A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

3 years ago

Can anyone help me with this question please

JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

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3 years ago