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3 years ago

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Factors that affect acceleration due to gravity.

2 answers:

hram777 [196]3 years ago

8 0

Answer

Factors Affecting Acceleration Due to Gravity

(1) Position on the planetary surface, it is higher at the axis (poles),

(2) Mass of the planet, higher with planets of larger masses

(3) Distance between object and the centre of planet, the larger the distance, the smaller the acceleration due to gravity.

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Kamila [148]3 years ago

5 0

Explanation:

the factors affecting acceleratiom due to gravity are:

Altitude above the earth’s surface.

Depth below the earth’s surface.

The shape of the earth.

Rotational motion of the earth.

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Two parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and w

Aleks [24]

Explanation:

The capacitance in two parallel-plate capacitors is:

$C=\frac{K\epsilon_oA}{d}$

For air, we have $K_1=1$

For plastic, we have $K_2=2.25$

Hence:

$C_1=\frac{K_1\epsilon_oA}{d_1}=\frac{\epsilon_oA}{d_1}\\C_2=\frac{K_2\epsilon_oA}{d_2}=2.25(\frac{\epsilon_oA}{d_2})$

a) Recall that the potential difference between the plates is the same ( $V_1=V_2=V$ ). The electric field is given by:

$E_1=\frac{V_1}{d_1}=\frac{V}{d_1}\\E_2=\frac{V_2}{d_2}=\frac{V}{d_1}$

The potential difference is defined as:

$V_1=\frac{Q_1}{C_1}=V\\V_2=\frac{Q_2}{C_2}=V$

Replacing:

$E_1=\frac{Q_1}{C_1d_1}\\E_2=\frac{Q_1}{C_1d_2}\\\\E_1d_1=\frac{Q_1}{C_1}\\E_2d_2=\frac{Q_1}{C_1}\\E_1d_1=E_2d_2\\\frac{E_1}{E_2}=\frac{d_2}{d_1}$

b) The energy is defined as:

$U=\frac{1}{2}CV^2$

So:

$U_1=\frac{1}{2}C_1V_1^2=\frac{1}{2}C_1V^2\\U_2=\frac{1}{2}C_2V_2^2=\frac{1}{2}C_2V^2\\U_1=\frac{1}{2}\frac{\epsilon_oA}{d_1}V^2\\U_2=\frac{1}{2}\frac{2.25\epsilon_oA}{d_2}V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=V^2\\\frac{2U_1d_1}{\epsilon_oA}=V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=\frac{2U_1d_1}{\epsilon_oA}\\\frac{U_1}{U_2}=0.44\frac{d_2}{d_1}$

5 0

3 years ago

A turntable has an angular velocity of 3.5 rad/s. A dust bunny is on the disk of the turntable at a distance of 0.2m from the ce

diamong [38]

Answer:

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off dust bunny from disk

m ω² r = μ m g

ω² r = μ g

3.5² x 0.2 = μ x 10 ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

3 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

$W=K_f -K_i$

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving,

$K_f = K_i + W=66,120+(-36,733)=29387 J$

The final kinetic energy of the car can be written as

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

where

m = 661 kg is its mass

v is its final speed

Solving for v,

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

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4 years ago

Based on the magnetic field lines shown what is the orientation of the mystery magnet ?

babunello [35]

I think it’s cannot be determined

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3 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

Andrews [41]

Explanation:

It is given that,

Mass of the runner, m = 70 kg

Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, $A=110\ mm^2=0.00011\ m^2$

Part A,

Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, $Y=0.15\times 10^{10}\ N/m^2$

The formula of the Young modulus is given by :

$Y=\dfrac{F/A}{\dfrac{\Delta L}{L}}$

$0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}$

$\Delta L=0.0049\ m$

Part B,

The fraction of the tendon's length does this correspond is given by :

$\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}$

$\dfrac{\Delta L}{L}=0.0326$

Hence, this is the required solution.

6 0

3 years ago