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Vedmedyk [2.9K]
3 years ago
8

Almost all jobs require that you fill out an _____.

Physics
2 answers:
zvonat [6]3 years ago
3 0

Almost all jobs require that you fill out an _____.;

D. Application

HACTEHA [7]3 years ago
3 0

Answer by JKismyhusbandbae: D. application

When you go to a job interview you fill out an application about all the jobs or things you've done called a resumé to see if you are fit for the job.

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
An ion has unequal numbers of which two particles?
Scrat [10]
Protons and electrons
7 0
3 years ago
What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and
inysia [295]

Answer:

25,000,000 Km ;)

Explanation:

5 0
3 years ago
a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m
Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

e = \frac{V-v}{u-U}

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0
3 years ago
Please help!!!!!
adoni [48]
Gpe is basses on the force equation...
GPE=m*g*h=1.5kg*9.8m/s^2*8m=117.6 N*m
8 0
2 years ago
Read 2 more answers
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