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2 years ago

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Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as great, the magnitude of the centr

ipetal force necessary to keep the car moving in the same path would be

1 answer:

kondaur [170]2 years ago

6 0

Answer:

We need 4 times more force to keep the car in circular motion if the velocity gets double.

Explanation:

Lets take the mass of the car = m

The radius of the arc = r

$F=\frac{m\times v^2}{r}$

Given that speed of the car gets double ,v' = 2 v

Then the force on the car = F'

$F'=\frac{m\times v'^2}{r}$ ( radius of the arc is constant)

$F'=\frac{m\times (2v)^2}{r}$

$F'=4\times \frac{m\times v^2}{r}$

We know that $F=\frac{m\times v^2}{r}$

Therefore F' = 4 F

So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.

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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0

1 year ago

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

olga55 [171]

Initially, the energies are:

$U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}$

At final point, the energies are:

$U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0$

Using conservation law of energy,

$-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}$

The equation is further simplified as:

$r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}$

7 0

1 year ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0

2 years ago

Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den

Ymorist [56]

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

4 0

2 years ago

Read 2 more answers

Kim has a metal casting company which makes commemorative coins. She has 0.12 cubic meters of silver which she needs to make int

a_sh-v [17]

Answer:

The coin has a diameter of 2.67 cm

Explanation:

First, we need to find the volume of each coin by dividing the total volume of silver by the number of coins. We have also to do a conversion of units in terms of centimeters as follows:

$V=0.12 m^3\times (\frac{100cm}{1m})^3=12000\ cm^3\\V_c=\frac{120000\ cm^3}{1.07\times10^5}= 1.121 cm^3$

Then, we define the coin as a tiny cilinder to determine its diameter. In that order we use the cilinder's volumen equation as follows:

$V=\pi r^2h\\r = \sqrt \frac{V}{\pi h}= \sqrt\frac{1.121 cm^3}{\pi \times 0.2cm}=1.336 cm$

Finally, we know that the diameter is twice the radius, therefore the diameter of each coin is 2.67 cm.

5 0

2 years ago