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3 years ago

6

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

t. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)

1 answer:

Mice21 [21]3 years ago

3 0

Answer:

r = 5.08 m

Explanation:

The electric force of attraction or repulsion is given by :

$F=\dfrac{kq_1q_2}{r^2}$

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

$\dfrac{kq_pq_e}{r^2}=mg$

r is distance

$r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m$

So, proton have to be at a distance of 5.08 meters above the electron.

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What is the velocity of a snail that travels 0.20 meters in 1 minute

stepladder [879]

The speed is 0.2 meter per minute.

There is not enough information given in the question to determine the velocity.

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3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

6 0

3 years ago

A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t

marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

864 = 240 t + 16 t²

Let's solve the second degree equation

16 t² + 240 t - 864 = 0

Let's divide by 16

t² + 15 t - 54 = 0

The solution of this equation is

t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

t = [-15 ±√(225 +216)] / 2

t = [-15 + - 21] / 2

We have two solutions.

t₁ = [-15 +21] / 2

t₁ = 3 s

t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0

3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

5 0

3 years ago

what is the difference between kinetic energy and random motion of a cylider of oxygen carried by a car and one standing on a pl

pochemuha

Answer:

See explanation below

Explanation:

If we are talking about the kinetic energy of the cylinder of oxygen:

The kinetic energy possessed by any object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this case, we have one cylinder carried by a car and one standing on a platform: this means that the speed of the cylinder carried by the car will be different from zero (and so also its kinetic energy will be different from zer), while the speed of the cylinder standing on the platform will be zero (and so its kinetic energy also zero). Therefore, the kinetic energy of the cylinder carried by the car will be larger than that standing on a platform.

Instead, if we are talking about the kinetic energy due to the random motion of the molecules of oxygen inside the cylinder:

The kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas:

$K=\frac{3}{2}kT$

where k is called Boltzmann constant and T is the absolute temperature of the gas. Therefore, we see that K does not depend on whether the gas is in motion or not, but only on its temperature - therefore, in this case there is no difference between the kinetic energy of the cylinder carried by the car and that standing on the platform (assuming they are at the same temperature)

6 0

3 years ago