Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?What is the electrostatic force between and electron and a proton separated by 0.1 mm?

Answers

Answer 1

Answer:

Answer:

The electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$

Explanation:

It is given that, charge $q_1$ is placed at a distance $r_o$ from charge $q_2$ . The force acting between charges is given by :

$F=(kq_1q_2)/(r_o^2)$

We need to find the force if the distance between them is reduced to $r_o/4$ . It is given by :

$F'=(kq_1q_2)/((r_o/4)^2)$

$F'=16* (kq_1q_2)/(r_o^2)$

$F'=16* F$

So, if the the distance between them is reduced to $r_o/4$ , the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or $10^(-4)\ m$ is :

$F=(kq_1q_2)/(r_o^2)$

$F=(9* 10^9* (1.6* 10^(-19))^2)/((10^(-4))^2)$

$F=2.30* 10^(-20)\ N$

So, the electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$ . Hence, this is the required solution.

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A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

Answers

Answer:

(a) $90^(\circ)$

(b) Initial velocity of the projectile is 22.54 m/s

(d) Parabolic

(e) The initial velocity is 23.04 m/s

Solution:

As per the question:

Velocity of the cart, v = 0.500 m/s

Distance moved by the cart, d = 2.30 m

Now,

(a) The projectile must be fired at an angle of $90^(\circ)$ so that it mounts on the top of the cart moving with constant velocity.

(b) Now, for initial velocity, u':

Time of flight is given by;

$T = (D)/(v)$ (1)

where

T = Flight time

D = Distance covered

(b) The component of velocity w.r.t an observer:

Horizontal component, $v_(x) = u'cos\theta$

Vertical component, $v_(y) = u'sin\theta - gT$

Also, the vertical component of velocity at maximum height is zero, $v_(y) = 0$

Therefore, $T = (u')/(g)$

Total flight time, $(2u')/(g)$ (2)

Now, from eqn (1) and (2):

$u' = (gD)/(2v)$

$u' = (9.8* 2.30)/(2* 0.500) = 22.54 m/s$

(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic

(e) The initial velocity is given by:

u = u' + v = 22.54 + 0.5 = 23.04 m/s

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answers

Answer:

a) W₁ = - 127 J, b) W₂ = 148.18 J, c) $v_(f)$ = 3.43 m/s and d) $v_(f)$ = 3.43 m / s

Explanation:

The work is given the equation

W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

W-T = m a

T = W -m a

T = mg -mg/7

T = mg 6/7

T = 3.6 9.8 6/7

T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

W₁ = F d cos θ

W₁ = 30.24 4.2 cos 180

W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

W₂ = (mg) d cos 0º

W₂ = 3.6 9.8 4.2

W₂ = 148.18 J

c) kinetic energy

K = ½ m v²

Let's calculate speed with kinematics

$v_(f)$ ² = vo² + 2 a y

v₀ = 0

a = g / 7

$v_(f)$ ² = 2g / 7 y

$v_(f)$ = √ (2 9.8 4.2 / 7)

$v_(f)$ = 3.43 m/s

We calculate

K = ½ 3.6 3.43²

K = 21.18 J

d) the speed of the block and we calculate it in the previous part

$v_(f)$ = 3.43 m / s

A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

Answers

Since volume of each cube is 1 cm^3
Then we can get the volume of 1 mole of cubes, which is $1 * 6.02 * 10^23 cm^3$
The edge $edge = v^1/3$
And the new adge that we are looking for: $new edge = (6.02*10^23)^1/3$ = $= 1.8191 * 46415888.336 = 84435142.472$
So, the final soution for the edge length of cube is 844km.

Do hope it helps!
Regards.

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floor and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?

Answers

Answer:

Force, $F=2.02* 10^6\ N$

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, $P=1\ atm=101325\ Pa$

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

$P=(F)/(A)$

$F=P* l* b$

$F=101325* 4* 5$

$F=2.02* 10^6\ N$

So, the total force on the floor due to the air above the surface is $2.02* 10^6\ N$ . Hence, this is the required solution.

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

Answers

Answer:

The linear velocity is represented by the following expression: $v = (s)/(t)$

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

$v = r\cdot \omega$ (Eq. 1)

Where:

$r$ - Radius of rotation of the particle, measured in meters.

$\omega$ - Angular velocity, measured in radians per second.

$v$ - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

$\omega = (\theta)/(t)$ (Eq. 2)

Where:

$\theta$ - Angular displacement, measured in radians.

$t$ - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

$v = (r\cdot \theta)/(t)$ (Eq. 3)

From Geometry we must remember that circular arc ( $s$ ), measured in meters, is represented by:

$s = r\cdot \theta$

$v = (s)/(t)$

The linear velocity is represented by the following expression: $v = (s)/(t)$

A cylinder with a diameter of 2.0 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?

Answers

Answer:

3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled

Explanation:

see the attachment

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