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4 years ago

The Milky Way is our Galax. Where are we located in Milky Way?

1 answer:

4 0

Earth is located in one of the spiral arms of the Milky Way (called the Orion Arm) which lies about two-thirds of the way out from the center of the Galaxy.

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A car of mass 1200Kilograms moving at 15 m/s the driver applies the brakes for 0.08 seconds and the castles down to 10 meter per

denpristay [2]

Initial velocity=u=15m/s
Final velocity=v=10m/s
Time=0.08s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-15}{0.08}$

$\\ \sf\longmapsto Acceleration=\dfrac{-5}{0.08}$

$\\ \sf\longmapsto Acceleration=-62.5m/s^2$

5 0

3 years ago

A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglec

Novosadov [1.4K]

Answer:

The free-body diagram of the cannonball is found in the attachment below

Note The question is incomplete. The complete question is as follows:

A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.

Draw the free-body diagram of the cannonball.

Explanation:

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.

In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:

F = force exerted by the cannon acting in the direction of angle of projection

Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon

Fw = weight of the cannonball acting in a downward direction

The free body diagram is as shown in the attachment below.

6 0

3 years ago

What happened to the speed of light if it travels from air into glass?

vovikov84 [41]

Answer:

it slows down

Explanation:

the answer

8 0

3 years ago

Find time when boy catches the girl or when they are at their closest separation..

soldier1979 [14.2K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\$

$\green{\underline{\bold{Given :}}} \\ \\ \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\ \\ \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\ \\ \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy = {1 \: m/s}^{2}} \\ \\ \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl= {2\: m/s}^{2} }\\ \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl }\\$

According to given question :

$\\ \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\\\ \\ \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\ \tt\purple{: \implies s = ut + \frac{1}{2} {at}^{2}} \\ \\ \tt\green{: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2}} \\ \\ \tt\purple{: \implies 300 = 40t - {t}^{2}} \\ \\ \tt\green{: \implies {t}^{2} - 40t + 300 = 0} \\ \\ \tt\purple{: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} }\\ \\ \tt\green{: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} }\\ \\ \tt\purple{: \implies t = \frac{40 \pm 20}{2} }\\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$

7 0

3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago

The Milky Way is our Galax. Where are we located in Milky Way?​

The Milky Way is our Galax. Where are we located in Milky Way?