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FromTheMoon [43]
2 years ago
6

Voltage can be provided by

Physics
1 answer:
kodGreya [7K]2 years ago
5 0
All of these can provide voltage. a battery is a generator, and plugging in to an electrical outlet is just like your phone charger or TV..
You might be interested in
A2 kg object moves at a constant 5 m/s across a level surface. Determine its kinetic energy.
AURORKA [14]
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
6 0
2 years ago
Draw the following vector quantity Using the coordinate system.
DiKsa [7]

The given vectors quantities can be described by their properties of both

magnitude and direction.

  • a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
  • b. The velocity vector extending from  (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
  • c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

\vec{F} = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; \vec{v} = 120 × cos(25)·i + 120×sin(25)·j, which gives;

\vec{v} = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

\vec{v} ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is \vec{d} = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

\vec{d} = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

brainly.com/question/10409036

4 0
2 years ago
In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in
aleksandrvk [35]

Answer:

A = 5.6μs

B = 178.57kHz

C = 2.8μs

Explanation:

A. It takes ¼ of the period of the circuit before the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor.

t = T/4

T = 4*t

T = 4 * 1.4 = 5.6μs

B. f = 1/T

Frequency is the inverse of period

f = 1 / 5.6*10⁻⁶

f = 178571.4286Hz

f = 178.57kHz

C. time taken for maximum energy to occur is T/2

t = 5.6 / 2 = 2.8μs

5 0
3 years ago
A 398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed
kicyunya [14]

Answer:

Explanation:

We shall represent speed in vector form

First speed

v₁ = 1.5 cos 14 i + 1.5 sin 14 j

= 1.455 i + 0.363 j

v₂ = 4.4 cos 33 i + 4.4 sin 33 j

= 3.69 i + 2.39 j

v₂ - v₁

3.69 i + 2.39 j - 1.455 i - 0.363 j

= 2.235 i + 2.027 j

acceleration

=  v₂ - v₁ / time

= ( 2.235 i + 2.027 j  ) / 23

= .097 i + .088 j

force = mass x acceleration

= 398 x ( .097 i + .088 j )

= 38.6 i + 35.02 j

Magnitude of force F

F² = 38.6² + 35.02²

F = 52.11 N

Tan θ = 35.02 / 38.6

θ = 42° north of east.

4 0
2 years ago
A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall
Alexandra [31]

Answers:

a) \theta_{2}=38\°

b) t=0.495 s

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{1} t_{1}   (1)

Where:

V_{o}=14.1 m/s is the initial speed  of snowball 1 (and snowball 2, as well)

\theta_{1}=52\° is the angle for snowball 1

t_{1} is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

y=0  is the final height of the  snowball 1

g=-9.8m/s^{2}  is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{2} t_{2}   (3)

Where:

\theta_{2} is the angle for snowball 2

t_{2} is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}   (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate t_{1} from (2):

0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (5)

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}   (6)

Substituting (6) in (1):

x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})   (7)

Rewritting (7) and knowing sin(2\theta)=sen\theta cos\theta:

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})   (8)

x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))   (9)

x=19.684 m   (10)  This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate t_{2} from (4) and substitute it on (3):

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}   (11)

x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})   (12)

Rewritting (12):

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})   (13)

Finding \theta_{2}:

2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})   (14)

2\theta_{2}=75.99\°  

\theta_{2}=37.99\° \approx 38\°  (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle (\theta_{2} < \theta_{1}).

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of t_{1} and t_{2} from (6) and (11), respectively:

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}  

t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}   (16)

t_{1}=2.267 s   (17)

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}  

t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}   (18)

t_{2}=1.771 s   (19)

Since snowball 1 was thrown before snowball 2, we have:

t_{1}-t=t_{2}   (20)

Finding the time difference t between both:

t=t_{1}-t_{2}   (21)

t=2.267 s - 1.771 s  

Finally:

t=0.495 s  

4 0
3 years ago
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