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3 years ago

STOP WHAT UR DOING!!! TAKE A MOMENT TO THINK ABOUT THESE!!

Physics

1 answer:

dolphi86 [110]3 years ago

4 0

Answer:

lol

can i have brainllest

Explanation:

Think of this how can a camera have a circler lens but take rectangular photos

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How is science controlled by the governerment

Vedmedyk [2.9K]

Answer:

the government's use and control of science. To clarify, I live in the U.S. I'm only interested in how the U.S. government dictates science. If you have any information or links that could educate me on this topic, I would be appreciative.

Explanation:

6 0

3 years ago

Read 2 more answers

What is the role of electrical forces in nuclear fission

77julia77 [94]

The electrical forces pulls nucleus apart

8 0

3 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic

Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with $a=2m/sec^2$ for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so $224=\left ( 1+7 \right )v$

v = 28 m/sec

4 0

3 years ago

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago