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frosja888
9 days ago
10

A 3 kg rubber block is resting on wet concrete. The coefficient of static friction is 0.3. What is the minimum force that must b

e applied to the block to make the block begin to accelerate?
Physics
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How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n
inna [3103]

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

7 0
3 months ago
1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the
kicyunya [3294]
The force exerted on the car during the stop measures 6975 N. Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
6 0
1 month ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
kicyunya [3294]

Result:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

v=\sqrt{k\frac{e^2}{m_e r}}

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

r=5.3\cdot 10^{-11}m

while the mass of the electron is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

By plugging in the values into the formula, we achieve

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
1 month ago
Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
2 months ago
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