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koban [17]
3 years ago
5

Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos

itive x-axis. Find the direction and magnitude of the resultant of these forces.
Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

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Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas
Westkost [7]

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

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3 years ago
PLEASE HELP I NEED THIS TO PASS THE EIGHTH GRADE AND I ONLY HAVE A COUPLE HOURS LEFT!!!!!!
zavuch27 [327]

Answer:

the pe at the top of the building: 784 J

the pe halfway through the fall: 392 J

the pe just before hitting the ground: 784 J

Explanation:

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7 0
3 years ago
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
Which graph illustrates constant speed and velocity?
boyakko [2]

The correct graph is <u>D</u>.

The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.

The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.

The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.

The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.

8 0
3 years ago
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A 5 N force is applied to a 3 kg ball to change its velocity from 9 m/s to 3 m/s. What is the impulse on the ball ?
nika2105 [10]
So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns
3 0
3 years ago
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