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nataly862011 [7]
4 years ago
7

Two bowling balls are rolling toward each other. One of the balls, ball A, is 12 kg traveling to the right at a velocity of 3 m/

s, while the other ball, ball B, is 4 kg traveling at a velocity of 1 m/s to the left. They collide with each other and ball A bounces back to the left at a velocity of 1.2 m/s. How much velocity does ball B now have traveling to the right?

Physics
1 answer:
sladkih [1.3K]4 years ago
3 0
Si here is my best guess, hope it helps and it’s right

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An example would be gravity
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A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn
xenn [34]

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal =  47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

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3 years ago
If it takes 100 N to get a 10 kg object to accelerate at 10m/s/s, how much force will it take to get a 20 kg object to accelerat
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Answer: C) 200 N

Explanation:

The force F is defined as:

F=m.a

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m=20 kg is the mass of the object

a=10 m/s^{2} is the acceleration

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F=(20 kg)(10 m/s^{2})

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A traffic-safety engineer is designing a deceleration lane. She is basing the length of
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The deceleration of the car is observed as 2 m/s^2.

<h3>What is deceleration?</h3>

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u = 126 km/h to or 35 m/s

t = 8.5 s

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2 years ago
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
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0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

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r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
3 years ago
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