A bicycle is a compound machine, true or false

Scenario 3Starting at rest, a 3 kg ball is dropped from the side of a bridge and strikes the ground below at 35m/s. What is the

Nonamiya [84]

The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.

<h3>How can the height of a dropped ball be determined?</h3>

Y = 1/2 g t 2, where y is the height above the ground, g = 9.8 m/s2, and t = 1.3 s, is the formula for problems like these. Any freely falling body with an initial velocity of zero meters per second can use this formula. figuring out how much y is.

A ball drops from the top of a building and picks up speed as it descends. Its speed is increasing by 10 m/s every second. What we refer to as motion with constant acceleration is, for example, a ball falling due to gravity.

The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.

To Learn more About potential energy, Refer:

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4 0

2 years ago

a body is dropped from a height of 240m. (i)How long does it take to reach the ground. (ii) What is the velocity with which it w

kramer

Answer:

7.0 s, 69 m/s

Explanation:

If we take down to be positive, then the time to reach the ground is:

x = x₀ + v₀ t + ½ at²

240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.0 seconds

The final velocity is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)

v = 69 m/s

5 0

3 years ago

The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on

Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0

4 years ago

Find the total electric charge of 2.5 kg of electrons. Express your answer using two significant figures.

zimovet [89]

Answer : The total electric charge of electrons is, $-4.4\times 10^{11}C$

Explanation:

Answer : The number of electrons transferred are, $4.68\times 10^{20}$

Explanation :

First we have to calculate the number of electrons.

Number of electrons = $\frac{\text{Total mass of electrons}}{\text{Mass of one electron}}$

Mass of 1 electron = $9.1\times 10^{-31}kg$

Total mass of electron = 2.5 kg

Number of electrons = $\frac{2.5kg}{9.1\times 10^{-31}kg}$

Number of electrons = $2.75\times 10^{30}$

Now we have to calculate the total electric charge of electrons.

Formula used :

$Q=ne\\\\n=\frac{Q}{e}$

where,

n = number of electrons transferred = $2.75\times 10^{30}$

Q = charge on electrons = ?

e = charge on 1 electron = $-1.602\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$2.75\times 10^{30}=\frac{Q}{-1.602\times 10^{-19}C}$

$Q=-4.4\times 10^{11}C$

Thus, the total electric charge of electrons is, $-4.4\times 10^{11}C$

4 0

4 years ago

If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0

Alex73 [517]

Answer:

a) v = 54.7m/s

b) v = (58 - 1.66a) m/s

c) t = 69.9 s

d) v = -58.0 m/s

Explanation:

Given;

The height equation of the arrow;

H = 58t - 0.83t^2

(a) Find the velocity of the arrow after two seconds. m/s;

The velocity of the arrow v can be given as dH/dt, the change in height per unit time.

v = dH/dt = 58 - 2(0.83t) ......1

At t = 2 seconds

v = dH/dt = 58 - 2(0.83×2)

v = 54.7m/s

(b) Find the velocity of the arrow when t = a. m/s

Substituting t = a into equation 1

v = 58 - 2(0.83×a)

v = (58 - 1.66a) m/s

(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s

the time when H = 0

Substituting H = 0, we have;

H = 58t - 0.83t^2 = 0

0.83t^2 = 58t

0.83t = 58

t = 58/0.83

t = 69.9 s

(d) With what velocity will the arrow hit the surface? m/s

from equation 1;

v = dH/dt = 58 - 2(0.83t)

Substituting t = 69.9s

v = 58 - 2(0.83×69.9)

v = -58.0 m/s

8 0

3 years ago