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Firlakuza [10]
3 years ago
6

Why do you think that if you drop two balls, exactly two balls rise on the other side at a similar speed? Why doesn’t one ball f

ly off quickly instead? Why don’t three balls fly off slowly?
Physics
1 answer:
Zolol [24]3 years ago
7 0
It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).

: Hope this helps :)
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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg
telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²  (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t²  (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t  (while the motor is running)

v = v0 + g · t  (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

Now let´s find the velocity reached in that time:

v = v0 + a · t

v = 28 m/s² ·14 s

v = 3.9 × 10² m/s

Now, let´s find the maximum height reached by the rocket using the equations of height and velocity after the motor stops:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Notice that now v0 and y0 will be the velocity and height reached while the rocket was being accelerated in the upward direction, respectively.

Let´s find at which time the rocket reaches its maximum height. With that time, we can calculate the max-height.

At the maximum height, the velocity of the rocket is zero, then:

v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

t = 40 s

After the motor stops, it takes the rocket 40 s s to reach the maximum height.

Using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m +  3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

4 0
3 years ago
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

6 0
4 years ago
How long does it take Larry to run 200 meters running at a speed of 21 m/s
andreev551 [17]

Answer:

9.52 seconds

Explanation:

t = d/v

t = 200 / 21

t = 9.52s (to 3 s.f.)

6 0
2 years ago
Read 2 more answers
How could you increase electric force using each factor
NNADVOKAT [17]

Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.

5 0
4 years ago
A stone is thrown horizontally at a speed of 10.0 m/s from the top of a cliff 139.4 m high.
Lesechka [4]

Answer:

a) 14.2sec

b) 1394m away if horizontal speed never changes

c) 9.8m/s

Explanation:

5 0
3 years ago
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