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3 years ago

Two forces (4n and 3n) pull to the left while a 12n force pulls to the right. What is the net force?

2 answers:

5 0

Call the positive direction the one that points to the right.

negative force <=== ===> positive force

Then the forces are +12n, -4n, and -3n.

The net force is their sum: (+12n - 4n - 3n) = +5 n .

(5 newtons, pointing to the right) .

Ahat [919]3 years ago

3 0

6 i believe because it makes sense too me

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Acceleration is calculated by taking the change in velocity and dividing by the time. What is the

netineya [11]

Answer:

m/s/s

Explanation:

Acceleration is change in velocity over time.

a = Δv / t

a = [m/s] / [s]

a = [m/s/s]

4 0

3 years ago

a projectile whose masss is 10g is fired directly upward from ground level with an initial velocity of 1000m/s neglecting the ef

sammy [17]

Answer:

<em>The speed of the projectile when it impacts the ground is 1000 m/s</em>

Explanation:

<u>Vertical Launch</u>

When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.

The speed when the object is moving upwards is given by

$v_f=v_o-g.t$

The time it takes to reach the maximum height is when vf=0, i.e.

$0=v_o-g.t$

solving for t

$\displaystyle t=\frac{v_o}{g}$

The maximum height reached is

$\displaystyle y=\frac{gt^2}{2}=\frac{v_o^2}{2g}$

Then, the object starts to fall. The object's height is given by

$\displaystyle y=\frac{v_o^2}{2g}-\frac{gt'^2}{2}$

where t' is the time the object has traveled downwards. The height will be 0 again when

$\displaystyle \frac{v_o^2}{2g}=\frac{gt'^2}{2}$

Solving for t'

$\displaystyle t'=\frac{v_o}{g}$

We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is

$v_f=g.t'=v_o$

Thus, the speed of the projectile when it impacts the ground is 1000 m/s

4 0

4 years ago

A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim

xxTIMURxx [149]

Answer:

Placing cells in series increases the

voltage in the circuit by 1.5 V for each

cell. Increasing the voltage increases the

brightness of the bulb

Explanation:

I. MAKING THE CONNECTION

How many terminals are located on the battery? 2

How many terminals are located on the bulb? 2

II. PLACING CELLS IN SERIES

What is the effect on the brightness of the bulb by increasing the number of cells?

The bulbs become brighter when increasing the number of cells.

What changes occur in the current in the circuit when increasing the number of cells?

Increasing the number of cells increases the current in the circuit.

What changes occur in the voltage in the circuit when increasing the number of cells?

Increasing the number of cells increases the voltage (for cells in series the voltage is additive).

What changes occur in the resistance in the circuit when increasing the number of cells?

The resistance is determined by the number of bulbs. The resistance in the circuit remains unchanged.

III. PLACING BULBS IN SERIES

What is the effect on the brightness of the bulbs by increasing the number of bulbs?

Increasing the number of bulbs decreases the brightness of the bulbs.

What changes occur in the resistance in the circuit as more bulbs are added?

The resistance increases. In a series circuit, adding bulbs increases the resistance in the circuit.

What changes occur in the current in the circuit as more bulbs are added?

Increasing the resistance decreases the current.

Observations on unscrewing one bulb. Explain your observations.

A complete circuit requires the electrons to move from the negative terminal of the battery to the

positive terminal. When one bulb is unscrewed it opens the circuit preventing a complete circuit and

the electrons cannot return to the battery.

IV. PLACING BULBS IN PARALLEL

Compare the brightness with two bulbs in parallel with two bulbs in series.

Two bulbs in parallel are brighter than two bulbs in series.

How does increasing the number of circuits (bulbs) change the current and resistance?

In a parallel circuit each bulb is in its own circuit. As bulbs are added the resistance in the circuit

decreases since each circuit is another pathway for electrons to move from one end of the circuit

4 0

4 years ago

Write any two different between work and power?

attashe74 [19]

Answer:

1. Work Is a type of Physical Activity

2. Power is Basically Having control of Society

Explanation:

3 0

3 years ago

Read 2 more answers

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

Read 2 more answers