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3 years ago

Two examples of a stopping motion

2 answers:

7 0

Add the ball going at a fast speed then slowing down and coming to a complete stop.

Then you will have a car going down a slope the the speed of the car will slow down to the minimum speed of 0.

So these two will be used as examples of stopping motion

Anton [14]3 years ago

5 0

Pause?Freeze?
Stop?
Halt?

You might be interested in

A rock drops from a height of 60 m. how long does it take for it to hit the ground

Svet_ta [14]

Here, Apply 2nd equation of the Kinematics :
S = ut + 1/2 at²
Here, s = 60 m
u = 0 [ free fall ]
a = 9.8 m/s² [ constant value for Earth system ]

Substitute their values,
60 = 0*t + 1/2 * 9.8 * t²
120 = 9.8t²
t² = 120 / 9.8
t = √12.24
t = 3.50 s

In short, Your Answer would be 3.50 seconds

Hope this helps!

7 0

3 years ago

How does the input distance of a single fixed pulley compare to the out- put distance?

ololo11 [35]

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

Explanation:

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

4 0

3 years ago

17V:=192)

BartSMP [9]

Answer:

THAT loooks hard123 -

Explanation:

v hard..... 1234=123

vhemistryyyy

4 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

a ball is thrown straight up it passes a 2.00 m high window 7.5 meters off the ground on its path up it takes 1.3 s to go past t

Svet_ta [14]

Let's say the velocity at the bottom of the window was "v."
s = v*t + ½at²
2 m = v * 1.3s - 4.9m/s² * (1.3s)² = v * 1.3s - 8.3 m
v = 10.3m / 1.3s = 7.9 m/s

Then the initial speed was
V = √(v² + 2as) = √(7.9m/s² + 2 * 9.8m/s² * 7.5m) = 14 m/s ◄ initial velocity
(after rounding to 2 digits from 14.5 m/s).

8 0

3 years ago