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3 years ago

9

A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower

how long was it in the air ?
2.1 s
0.5 s
10 s
5 s

1 answer:

scoray [572]3 years ago

8 0

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

$d = 50 m$

And that the initial speed of the ball is

$u=25 m/s$

at an angle of

$\theta=20^{\circ}$

So, the horizontal speed of the ball (which is constant during the entire motion) is

$u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s$

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

$t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s$

which is the time the ball spent in air.

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2 years ago

Danielle exerts a 14.0 N force to compress a spring by a distance of 8.00 cm. What is the spring constant of this spring

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Answer:

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Explanation:

Given:

Force = F= 14.0 N

Distance = x = 8.00 cm = 0.08 m

To find:

spring constant

Solution:

spring constant is calculated by using Hooke's law:

k = F/x

Putting the values in above formula:

k = 14.0 / 0.08

k = 175 N/m

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3 years ago

What is the currency for performing work and what is its unit?

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The currency for performing work is called : Energy

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6 0

3 years ago

A 10 cm long tensile specimen with a yield stress of 350 MPa is pull to a stress of 300 MPa and then the load is released. What

Advocard [28]

Answer:

The length of the specimen after the load is released is 11.67 cm

Explanation:

Given;

yield stress, Y = 350 MPa

ultimate tensile stress, T = 300 MPa

Elongation factor, e = yield stress, Y / ultimate tensile stress, T

Elongation factor, e = 350 Mpa / 300 Mpa

Elongation factor, e = 1.1667

New length of the specimen = 1.1667 x 10 cm = 11.67 cm

Therefore, when the load is released from 10 cm long tensile specimen, the length of the specimen becomes 11.67 cm

3 0

3 years ago

Sam, whose mass is 72 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 190 N and a coefficien

Westkost [7]

Answer:

<em>181.11 m</em>

Explanation:

Net Force is the sum of all the resultant force of all the forces acting on a body. The net force experienced by the jet is the difference of the Applied force and frictional force and this is represented in equation 1;

$F_{net} = F_{applied} - F_{friction} \\$ ............................1

but $F_{friction}$ = μmg .......................................2

Where μ is the coefficient of friction = 0.1

m is the mass of the body = 72 kg

g is the acceleration due to gravity 9.81 m/ $s^{2}$

Substituting into equation 2 we have;

$F_{friction}$ = 0.1 x 72 x 9.81

$F_{friction}$ = 70.63 N

Now we substitute our answer in equation 1;

$F_{net} = 190 N - 70.63 N \\$

$F_{net}$ = 119.37 N

We have to calculate the net acceleration in order to get our velocity .

$F_{net}$ = m $a_{net}$

$a_{net}$ = $\frac{F_{net}}{m}$

$a_{net}$ = $\frac{119.37 N}{72 kg }$

$a_{net}$ = 1.66 $m/s^{2}$

Speed can be express as;

v = u + at ...................3

where u is the initial velocity, which is 0 in this case.

a is the net acceleration = 1.66 $m/s^{2}$

t is the time which is 9 s

substituting the values in equation 3 we have

v = 1.66 $m/s^{2}$ x 9 s

v = 14.94 m/s

Calculating for Sam's distance for the first 9 seconds, using the equation of motion we have;

$S_{1} = ut +\frac{1}{2}at^{2}$

the jet was initially at rest so initial velocity is 0 and $a_{net}$ = 1.66 $m/s^{2}$

$S_{1} = \frac{1}{2} *1.66 m/s^{2} *9^{2}$

$S_{1}$ = 67.23 m

Also we have to calculate Sam's distance after nine seconds using equations of motion express below;

$v^{2} -u^{2} = 2as$

making S the subject formula we have;

$S_{2} =\frac{v^{2}-u^{2} }{2a}$ ....................................4

v is the maximum velocity after the fuel finished and its 14.94 m/s and a is the acceleration along the horizontal plane which put into consideration the coefficient of friction. a = μg = 0.1*9.8 m/ $s^{2}$ = 0.98 m/ $s^{2}$

We substitute our values into equation 4 to get our remaining distance;

$S_{2} = \frac{(14.94 m/s)^{2} }{2(0.98 m/s^{2} )}$

$S_{2}$ = 113.88 m

Therefore the total distance S = $S_{1}$ + $S_{2}$

S = 67.23 m + 113.88 m

<em>The total distance covered by Sam is </em>181.11 m

8 0

3 years ago

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