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Fiesta28 [93]
3 years ago
5

The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10

−18 JJ of kinetic energy in this distance?

Physics
2 answers:
Serjik [45]3 years ago
8 0

Answer:

a=1.1*10^{18}\frac{m}{s^2}

Explanation:

We use the following kinematic formula to calculate the acceleration:

v_f^2=v_0^2+2ax

The kinetic energy is defined as:

\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}

Replacing this in the acceleration formula and solving for a:

\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}

Ilya [14]3 years ago
4 0

The solution is in the attachment

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What is the increase in pressure required to decrease volume of mercury by 0.001%
REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0
3 years ago
What is the direction of the net force that acts on an object undergoing uniform circular motion?
8_murik_8 [283]
D. The direction of the force is toward the center of the object's circular path.
5 0
3 years ago
Read 2 more answers
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

4 0
3 years ago
Acceleration toward the center of a curved path is called
Serggg [28]

Answer:

Centripetal acceleration.

Explanation:

Centripetal acceleration is a property of a body moving in a uniform circular path and it is directed radially towards the center of the circle in which body is rotating.

The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.

It is calculated through following formula

a=v^2/r

where v is velocity and r is the radius of the circle.

7 0
3 years ago
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
4 years ago
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