Three ways we can use an electric current

How do solve for a in yf=yi+vi *t + 1/2*at^2. Also, What’s this formulas name

ddd [48]

Answer:

Kinematics

Explanation:

8 0

4 years ago

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A solid sphere of radius R is made of an insulating material. It holds a charge, Q, which is distributed evenly throughout the s

Vinvika [58]

Answer:

The magnitude of Electric Field is $E=\dfrac{Qr}{4\pi \epsilon_0 R^3}$

Explanation:

Given:

Radius of the solid sphere=R
Total charge of the sphere=Q

Let consider a Gaussian surface at a distance of r such that 0<r>R in the shape of sphere such that the electric Field due to this E and it is radially outwards.

The charge inside this Gaussian surface volume we have , $q_{in}=\dfrac{Qr^3}{R^3}$

Now using Gauss Law we have

$E\times4\pi r^2=\dfrac{q_{in}}{\epsilon_0}\\E\times4\pi r^2=\dfrac{\dfrac{Qr^3}{R^3}}{\epsilon_0}\\E=\dfrac{Qr}{4\pi \epsilon_0 R^3}$

5 0

3 years ago

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i

Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

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

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

$v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s$

(b) 4.72 s

The vertical position of the rock at time t is given by

$y(t) = v_y t - \frac{1}{2}gt^2$

where

$v_y = 21 m/s$ is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

$-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0$

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

7 0

4 years ago

Please i really need help! :/

valina [46]

Answer: D

Explanation:
Let us examine the given actions to see which ones generate heat and sound energy from mechanical energy.

A) Stretching a string.
The mechanical stretching creates tension in the string, which is released when the tension is removed. The generation of thermal or sound energy is minimal or negligible.

B) Squeezing a sponge ball
The sponge ball experiences compressive loading. This generates minimal or no heat and sound energy.

C) Throwing a ball upwards in the air
Air friction generates minimal or no heat at low velocities. At low velocities the pressure waves are too small to generate sound.

D) Striking a hammer on a nail.
A tremendous amount of force is applied over a small area to generate very high stresses that are in the plastic zone. A high amount of thermal energy is generated and the localized disturbance of the air generates audible sound.
This is the correct situation.

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3 years ago

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc

Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t² (when the rocket is accelerated upward).

y = y0 + v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t (when the rocket has fuel)

v = v0 + g · t (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 + v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0

4 years ago