Do the data for the first part of the experiment support or

When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde

Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

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

r is the separation between charges

$F\propto \dfrac{1}{r^2}$

$r=\sqrt{\dfrac{1}{F}}$

If F'= 2F

$r'=\dfrac{1}{\sqrt{2F} }$

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. $r'=\dfrac{1}{\sqrt{2F} }$ . Hence, this is the required solution.

6 0

3 years ago

A helicopter carries a 1000-kg-car suspended from a rope below it. The helicopter flies horizontally at a constant speed of 25 m

kolbaska11 [484]

Answer:

(a) Tension, T = 28.653 kN

(b) Wind resistance force, $26.925\ kN$

Solution:

As per the question:

Mass of the car, m = 1000 kg

Speed of the helicopter, v = 25 m/s

Angle made by the rope, $theta = 20^{\circ}$

Now,

(a) To calculate the tension, T in the car:

Tension along the direction of motion, $T_{h} = Tcos20^{\circ}$

Tension along the vertical direction, $T_{v} = Tsin20^{\circ}$

Now, let the force due to the wind directed in the opposite direction of the motion be $F_{W}$ and it balances the horizontal component of the tension, T.

The vertical component is balance by the weight of the car, i.e., mg that acts vertically downwards.

Now,

$T_{v} = mg$

$Tsin20^{\circ} = 1000\times 9.8$

T = 28653 N = 28.653 kN

(b) The force of the wind resistance:

$F_{W} = T_{h}$

$F_{W} = 2cos20^{\circ} = 26925\ N = 26.925\ kN$

(c) Now,

If the angle made by the rope with the vertical is $0^{\circ}$ :

$mg = Tsin(90^{\circ} - 0^{\circ})$

$Tsin90^{\circ} = mg = 9800\ N$

The tension in the rope will be equal to the weight the car.

Wind resistance force, $F_{W} = Tcos90^{\circ} = 0\ N$

If the angle made by the rope with the vertical is $90^{\circ}$ :

$mg = Tsin(90^{\circ} - 90^{\circ})$

T = 0 N

Wind resistance force, $F_{W} = Tsin0^{\circ}$

$Tsin0^{\circ} = mg$

$F_{W} = \infty$

There will be no tension in the rope and wind resistance will be infinite.

3 0

3 years ago

A rocket is launched straight up from the earth's surface at a speed of 1.80×104 m/s .part awhat is its speed when it is very fa

balandron [24]

We can solve the problem by using the law of conservation of energy.

When the rocket starts its motion from the Earth surface, its mechanical energy is sum of kinetic energy and gravitational potential energy:
$E_i = K_i + U_i = \frac{1}{2} m v_i^2 + (- \frac{GM}{r} )$
where
m is the rocket's mass
$v_i = 1.8 \cdot 10^4 m/s$ is the rocket initial speed
$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant
$M=5.97 \cdot 10^{24} kg$ is the Earth's mass
$r= 6.37 \cdot 10^6 m$ is the distance of the rocket from the Earth's center (so, it corresponds to the Earth's radius)

The mechanical energy of the rocket when it is very far from the Earth is just kinetic energy (because the gravitational potential at infinite distance from Earth is taken to be zero):
$E_f = K_f = \frac{1}{2} mv_f ^2$
where $v_f$ is the final speed of the rocket.

By equalizing the initial energy and the final energy, we can find the final velocity:
$\frac{1}{2} mv_i ^2 - \frac{GM}{r} = \frac{1}{2}m v_f^2$
$v_f = \sqrt{v_i^2 - \frac{GM}{r} } =1.41 \cdot 10^4 m/s$

3 0

4 years ago

Read 2 more answers

What will happen to the absorption and/or emission spectral lines of an object moving away from Earth at high speed?

Rufina [12.5K]

Answer:

c. They will be shifted toward the red end of the spectrum.

Explanation:

When the absorption or emission lines of a moving object away from from the earth at very high speed then the spectral lines will be shifted toward the red end of the spectrum.This effect can be easily understand by Doppler effect..

So the our option c is correct.

c. They will be shifted toward the red end of the spectrum.

4 0

3 years ago

which cell structure is responsible for taking in and breaking down foreign particles and damaged organelles?

Klio2033 [76]

Answer:

lysosomes

Explanation:

4 0

3 years ago