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4 years ago

A ball is dropped from a height of 2 m. how long will it take this ball to reach the ground?

Physics

1 answer:

alex41 [277]4 years ago

7 0

Answer:

It depends on the weight or mass or how big the ball is.

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Two waiters are trying to get through a single door of a kitchen. One pushes on one side of a door 0.567 m from the hinge with a

NISA [10]

Answer:

$275.5 N$

Explanation:

$F_{1}$ = Force on one side of the door by first waiter = 257 N

$F_{2}$ = Force on other side of the door by second waiter

$r_{1}$ = distance of first force by first waiter from hinge = 0.567 m

$r_{2}$ = distance of second force by second waiter from hinge = 0.529 m

Since the door does not move. hence the door is in equilibrium

Using equilibrium of torque by force applied by each waiter

$r_{1} F_{1} = r_{2} F_{2} \\(0.567) (257) = (0.529) F_{2}\\F_{2} = 275.5 N$

7 0

3 years ago

6. A ball rolled 125 m from the top of a hill to the bottom of a hill. How long

leva [86]

Time = distance / speed
T = 125/ 5
T = 25 meters per second

7 0

3 years ago

D=1/2at^2 <br> solve for a

nignag [31]

Answer:

a = 2d / t²

Explanation:

d = ½ at²

Multiply both sides by 2:

2d = at²

Divide both sides by t²:

a = 2d / t²

4 0

3 years ago

Which best describes electrical energy? a. energy released by a chemical reaction b. energy produced by flow of electric charge

inessss [21]

<span> energy produced by flow of electric charge describes a electrical energy
because movement of electric charge do effect the work on system
so correct option is B
hope it helps</span>

4 0

3 years ago

Read 2 more answers

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 0

3 years ago