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3 years ago

How to solve these two questions ?

Physics

1 answer:

murzikaleks [220]3 years ago

8 0

1) See attached graph

To solve this part of the problem, we have to keep in mind the relationship between current and charge:

$i = \frac{\Delta Q}{\Delta t}$

where

i is the current

Q is the charge

t is the time

The equation then means that the current is the rate of change of charge over time.

Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.

Here we have:

- Between t = 0 and t = 2 s, the slope is $\frac{50-0}{2-0}=25 C/s$ , so the current is 25 A

- Between t = 2 s and t = 6 s, the slope is $\frac{-50-(50)}{6-2}=-25 C/s$ , so the current is -25 A

- Between t = 6 s and t = 8 s, the slope is $\frac{0-(-50)}{8-6}=25 C/s$ , so the current is 25 A

Plotting on a graph, we find the graph in attachment.

2) $15 \mu C$

The relationship we have written before

$i = \frac{\Delta Q}{\Delta t}$

Can be rewritten as

$\Delta Q = i \Delta t$

This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.

Therefore, we need to find the area under the graph.

Here we have a trapezium, where the two bases are

A = 1 ms = 0.001 s

B = 2 ms = 0.002 s

And the height is

h = 10 mA = 0.010 A

So, the area is

$Area=\frac{(0.001+0.002)\cdot 0.010}{2}=1.5\cdot 10^{-5} C = 15 \mu C$

So, the charge is $15 \mu C$ .

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Answer:

Since the five forces applied are all in the same direction with the motion of the sled, to solve for the net force, add the five forces together. Therefore, the net force is 30 Newtons.

Explanation:

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3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height

Svetllana [295]

Answer:

Part A:

(a): -121.26 ft/s.

(b): -121.13 ft/s.

(c): -121.052 ft/s.

(d): -121.026 ft/s.

Part B:

-121.00 ft/s.

Explanation:

Given that the height of the balloon after t seconds is

$\rm y(t) = 35 t-26t^2.$

The average velocity of an object is defined as the total distance traveled by the object divided by the time taken in covering that distance.

$\rm v_{av} = \dfrac{y_2-y_1}{t_2-t_1}$

where,

$\rm y_2,\ y_1$ are the positions of the object at time $\rm t_1$ and $\rm t_2$ respectively.

For the average velocity for the time period beginning when t=3 and lasting .01 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.01\ sec = 3.01\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.01-26\cdot 3.01^2=-130.2126\ ft.\\\\\Rightarrow v_{av}=\dfrac{-130.2126-(-129)}{3.01-3}=-121.26\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .005 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.005\ sec = 3.005\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.005-26\cdot 3.005^2=-129.60565\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.60565-(-129)}{3.005-3}=-121.13\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .002 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.002\ sec = 3.002\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.002-26\cdot 3.002^2=-129.2421\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.2421-(-129)}{3.002-3}=-121.052\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .001 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.001\ sec = 3.001\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.001-26\cdot 3.001^2=-129.121\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.121-(-129)}{3.001-3}=-121.026\ ft/s.$

The instantaneous velocity of the balloon at the given time is defined as the rate of change of its position at that time.

$\rm v(t) = \dfrac{dy}{dt}\\=\dfrac{d}{dt}\left ( 35 t-26t^2\right )\\\\=35-26\times 2t.\\\\At\ t=3,\\\\v(t)=35-26\times 2\times 3=-121.00\ ft/s.$

<u>Note:</u><em> The negative sign with all the velocities indicates that the direction of these velocities are downwards.</em>

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3 years ago

A circular loop of radius 13 cm carries a current of 13 A. A flat coil of radius 0.94 cm, having 58 turns and a current of 1.9 A

telo118 [61]

Answer:

$6.28\times 10^{-5}\ \text{T}$

$1.92\times 10^{-6}\ \text{Nm}$

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi 10^{-7}\ \text{H/m}$

$I_l$ = Current in circular loop = 13 A

$r_l$ = Radius of circular loop = 13 cm

$N$ = Number of turns = 58

$r_c$ = Radius of coil = 0.94 cm

$I_c$ = Current in coil = 1.9 A

$\theta$ = Angle between loop and coil = $90^{\circ}$

Magnitude of magnetic field in circular loop

$B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}$

The magnetic field produced by the loop at its center is $6.28\times 10^{-5}\ \text{T}$ .

Torque is given by

$\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}$

The torque on the coil due to the loop $1.92\times 10^{-6}\ \text{Nm}$ .

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2 years ago

A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down

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The maximum is 1800 J since energy is not created nor destroyed, it is just transformed. In this example, gravitational potential energy is transformed to kinetic energy.

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2 years ago

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Answer:

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How to solve these two questions ?​

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