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STALIN [3.7K]
3 years ago
8

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?
Physics
1 answer:
ANEK [815]3 years ago
8 0
The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
F=qE
where in our problem q=3 \mu C= 3 \cdot 10^{-6} C and E=2 kN/C=2000 N/C, so the force is
F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J
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Please Awnser soon water's state of matters include stream, liquid water, and ice. what about water is the same in these states?
yulyashka [42]
Well the similarity is that even though they are in a different state of matter they still come from the same substance: h2o 
7 0
3 years ago
A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu
neonofarm [45]

Answer:

<h2> 48kg</h2>

f = ma \\

w \:  =  \frac{f}{g}  \\

Explanation:

f \:  = ma \\ f - mg = ma \\ f = ma + mg \\ f = 40 \times 2 + 40 \times 10 \\ f = 480

w =  \frac{f}{g}  \\w =   \frac{480}{10}  \\ w = 48kg

5 0
3 years ago
An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi
Lynna [10]

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

6 0
3 years ago
Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.
ivann1987 [24]

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

3 0
3 years ago
The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0
3 years ago
Read 2 more answers
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