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andreev551 [17]
4 years ago
15

A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air r

esistance.]
Physics
2 answers:
Elza [17]4 years ago
6 0
<span>2.856s Distance traveled under constant acceleration as a function of time is given by x=(gt^2)/2 where g is the acceleration and t is time. In this case acceleration is due to gravity and is 9.81m/s^2. The distance of interest is x=40m. Substituting and solving 40=(9.81*t^2 )/ 2 80 = 9.81*t^2 8.1549 = t^2 2.856s = t</span>
Effectus [21]4 years ago
3 0
H = 40 m, the height from which the ball is dropped.
m = 1 kg, the mass of the ball

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

The initial vertical velocity is zero.
If t = the time of flight, then
40 m = (1/2)*g*(t s)² = 0.5*9.8*t²
t² = 40/4.9 = 8.1633
t = 2.857 s

Answer:  2.9 s (nearest tenth)
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Answer:

Explanation:

It 100 kg It think

6 0
3 years ago
What is the acceleration of a 150 kg object pushed with a force of 1000 N?
kirill115 [55]

Answer:

<h2>6.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{1000}{150}   \\  =  6.66666..

We have the final answer as

<h3>6.67 m/s²</h3>

Hope this helps you

8 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
1.
MAXImum [283]

Answer:

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Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles

7 0
3 years ago
The closest star to our solar system is Alpha Centauri, which is 4.12 × 10^16 m away. How long would it take light from Alpha Ce
bonufazy [111]

Time = (distance) / (speed)

Time = (4.12x10^16 m) / (3 x10^8 m/s)

Time = 1.37 x 10^8 seconds

Divide the seconds by 86,400 to get days. Then divide the days by 365 to get years.

Time = about 4.35 years

4 0
3 years ago
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