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2 years ago

A 1 mg ball carrying a charge of 2 x 10-8 C hangs from a

Physics

1 answer:

Fed [463]2 years ago

7 0

Answer:

σ = 0.255*10^-3 C/m²

Explanation:

The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.

At equilibrium ,

Upward force = downward force

Tcosθ = mg ----(1)

Assuming that the Forward force = backward force, then

Tsinθ = σq/ε₀

[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)

Dividing equation (2) by (1)

Tsinθ/Tcosθ = qσ/ε₀mg

⇒Tanθ = qσ/ε₀mg

σ = ε₀mg tanθ/q

Now substituting the values of

σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8

σ = (8.85*10^-12 * 0.5774) / 2*10^-8

σ = 5.11*10^-12 / 2*10^-8

σ = 0.255*10^-3 C/m²

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A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t

Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m$

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s$

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

$v=u+at\\\Rightarrow v=0+9.81\times 86.47 = 848.27\ m/s$

Speed just before hitting the water is 848.27 m/s

3 0

2 years ago

A uniformly charged rod (length = 2.0 m, charge per unit length = 5.0 nc/m) is bent to form one quadrant of a circle. what is th

Neko [114]

Electric field at the center of circular arc is given by the formula

$E = \frac{2k\lambda sin\frac{\theta}{2}}{R}$

here we know that

$\lambda = 5nC/m$

$k = 9 \times 10^9 Nm^2/C^2$

$\frac{\pi}{2}R = L = 2m$

$R = \frac{4}{\pi}$

also we know that

$\theta = \frac{\pi}{2}$

now from above formula

$E = \frac{2(9\times 10^{-9})(5\times 10^{-9})sin45}{4/\pi}$

$E = 50 N/C$

4 0

2 years ago

A free falling object has the velocity time graph shown. What is the objects displacement between 0.0 and 6.0s

zloy xaker [14]

For free fall motion the displacement can be found by graphically as well as by kinematics equation

Here acceleration of object is constant as it fall due to gravity so we can use

$d = v_i * t + \frac{1}{2}at^2$

here if body starts with zero initial speed then we can say

$d = 0 + \frac{1}{2}*9.8*t^2$

here we need to find the displacement from t = 0 to t = 6s

so we can say

$d = \frac{1}{2}*9.8*6^2$

$d = 176.4 m$

so the displacement will be 176.4 m

in order to find the displacement from the graph of velocity and time we need to find the area under the graph for given time interval that will also give us same displacement for given period of time.

6 0

2 years ago

Read 2 more answers

A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

inysia [295]

Hello!

F = Bqv sin theta F=Force B=magnetic flux density q=charge v=velocity theta=angle the moving electrons make with the magnetic field.

^^^You will find the force using that formula^^^

In Short, your Answer would MOST LIKELY have to be "B".

"<span>-3.9 × 10-14 N"
</span>
<span>I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)</span>

6 0

2 years ago

Read 2 more answers

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$