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Harrizon [31]
2 years ago
11

a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the

time when them cross each other
Physics
1 answer:
aleksklad [387]2 years ago
5 0
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

             H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

             H  =  20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

                         <span>50 - (1/2 G Q²)  =  20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add  </span><span>(1/2 G Q²)  to each side.
Then it says
                                             50  =  20Q

Divide each side by 20:          2.5  =  Q .

And there we are.  The stones pass each other

                                       2.5 seconds

after they are simultaneously launched.
</span>
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A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

8 0
3 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
3 years ago
A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app
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Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

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8 0
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Andrews [41]
First, we convert kcal to joules:
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475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

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ΔU = -500 + 1987.4
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4 0
3 years ago
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Answer:

F= 134.92 N

Explanation:

Given that

The mass of the moon ,M = 7.4 x 10²² kg

The mass of the man ,m = 79 kg

The radius ,R= 1.7 x 10⁶ m

The force exerted by moon is given as

F=G\dfrac{Mm}{R^2}

Now by putting the values in the above equation we get

F=6.67\times 10^{-11}\times \dfrac{79\times 7.4\times 10^{22}}{(1.7\times 10^6)^2}\ N\\F=134.92 N

Therefore the force will be 134.92 N.

F= 134.92 N

8 0
3 years ago
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