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3 years ago

A bicycle rides in a straight line at 1 km at 10

Physics

1 answer:

ahrayia [7]3 years ago

6 0

Answer:

avg speed =total distance/total time

total distance =2km (given)

total time=1/10 +1/25=7/50

so avg speed =100/7=14.28

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient

Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0

3 years ago

What influences the strength of an electric field?

Slav-nsk [51]

Answer:

Explanation:

The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.

4 0

3 years ago

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we

Alex_Xolod [135]

Answer:

w = 2w₀ the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

L₀ = L

I₀ w₀ = I w

I₀ w₀ = I₀ / 2 w

w = 2w₀

therefore the angular velocity of man doubles

7 0

3 years ago

2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n

AleksandrR [38]

Answer:

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

$d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}$