All categories

2 years ago

8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's

Physics

1 answer:

Dafna1 [17]2 years ago

5 0

Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

$k\frac{q_1 q_2}{r^2}=mg$

where

k is the Coulomb constant

$q_1=-1\cdot 10^{-10}C$ is the charge on the balloon

$q_2=-1\cdot 10^{-5} C$ is the charge on the bag

r is the separation betwen the bag and the balloon

$m=0.02 g=2\cdot 10^{-5} kg$ is the mass of the bag

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

Solving for r, we find the distance at which the bag must be held:

$r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m$

You might be interested in

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

Fofino [41]

Answer:

Force will be acting southward and the magnitude of force will be 1000 N

Explanation:

given,

mass of car = 1000 Kg

initial speed of the car (u) = 20 m/s

final speed of the car (v) = 0 m/s

distance to stop the car = 200 m

using equation of motion

v² = u² + 2 a s

0 = 20² + 2 x a x 200

400 a = -400

a = -1 m/s²

Now, we know

Force = mass x acceleration

F = 1000 x -1

F = -1000 N

- ve sign of force represent force will be acting in the opposite direction of motion.

Force will be acting southward and the magnitude of force will be 1000 N

6 0

2 years ago

An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars

Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

$W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2$

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

$W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}$

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

$P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}$

the car did not exceed the speed limit

Hope this helps!!

7 0

2 years ago

Read 2 more answers

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

2 years ago

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie

Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0

2 years ago

Read 2 more answers

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 mete

Verdich [7]

Answer:

<em>The glider's new speed is 68.90 m/s</em>

Explanation:

<u>Principle Of Conservation Of Mechanical Energy</u>

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

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

The gravitational potential energy when it's at a height h from the zero reference is

$U=mgh$

The total mechanical energy is

$M=K+U$

$\displaystyle M=\frac{1}{2}mv^2+mgh$

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

$\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)$