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statuscvo [17]
1 year ago
9

Baseballs pitched by a machine have a horizontal velocity of 30 meters/second. The machine accelerates the baseball from 0 meter

s/second to 30 meters/second in 0.5 seconds. If a baseball has a mass of 0.15 kilograms, the force the machine exerts is _______ newtons. Use F = ma, where a= v-u/t.
Physics
1 answer:
miv72 [106K]1 year ago
4 0

using the formula

F = ma

Where F is the force applied by the machine

A is the acceleration which is also equal to v/t where v is the velocity and t is time

M is the mass

 

F = mv/t

F = (0.15kg) (30 – 0 m/s)/ 0.5 s

<span>F = 9 N</span>

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A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the
nekit [7.7K]
The work done by the force pushing the puck is equal to the product between the force and the distance covered:
W=Fd=(750 N)(2.0 m)=1500 J

And the power is the work done per unit time:
P= \frac{W}{t}
and using t=0.30 s, we find the power:
P= \frac{1500 J}{0.30 s}=5000 W
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2 years ago
a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
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Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer:  11.025 m
7 0
2 years ago
Read 2 more answers
Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

4 0
1 year ago
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P
Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0
2 years ago
The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find
natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

F_A=G\frac{M_Am}{R_A^2}

For planet B,

F_B=G\frac{M_Bm}{R_B^2}

We also know that the weight of the object on the two planets is the same, so

F_A = F_B

So we can write

G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}

We also know that the mass of planet A is only sixty percent that of planet B, so

M_A = 0.60 M_B

Substituting,

G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}

Now we can elimanate G, MB and m from the equation, and we get

\frac{0.60}{R_A^2}=\frac{1}{R_B^2}

So the ratio between the radii of the two planets is

\frac{R_A}{R_B}=\sqrt{0.60}=0.775

6 0
1 year ago
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