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2 years ago

Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction

Physics

1 answer:

SpyIntel [72]2 years ago

3 0

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

$F = \frac{kq_1 q_2}{r^2}$

here we have

$q_1 = q_2 = 6 \times 10^{-6} C$

distance between two charges is given as

r = 0.50 m

so we have

$F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}$

$F = 1.296 N$

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A 4-kg mass moving with speed 2 m/s, and a 2-kg mass moving with a speed of 4 m/s, are gliding over a horizontal frictionless su

Akimi4 [234]

Answer:

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

Explanation:

As we know that both mass have same horizontal force opposite to their motion

So we will have

$a = \frac{F}{m}$

$a_1 = \frac{F}{4}$

$a_2 = \frac{F}{2}$

now the stopping distance of an object moving with initial speed v is given as

$v_f^2 - v_i^2 = 2(-a) d$

$d = \frac{v^2}{2a}$

so here we have

$d_1 = \frac{2^2}{\frac{F}{4}}$

$d_1 = \frac{16}{F}$

for other object we have

$d_2 = \frac{4^2}{\frac{F}{2}}$

$d_2 = \frac{32}{F}$

So correct answer will be

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

3 0

2 years ago

A bowling ball and a bag of potato chips are dropped at the same time from a tall tower on the moon. there is no air on the moon

rusak2 [61]

Both objects will hit the ground at the same time

8 0

3 years ago

Read 2 more answers

1. In physical science, work occurs when a ___

11111nata11111 [884]

Pretty sure it’s force so C

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8 0

1 year ago

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

5 0

2 years ago

Two sheets of polarizing material are placed such that their polarizing axes are 90° to each other and no light passes through t

nekit [7.7K]

Answer:

Explanation:

Polarization In this case angle of incidence is not equal to angle of polarization, hence reflected light is partially polarized and transmitted light is also partially polarized. by reflection is explained by Brewster's law,

According to this when unpolarized light incident on glass plate at an angle is called as angle of polarizing the reflected light is plane polarized, and transmitted light is partially polarized. The plane of vibration of polarized light is having plane of vibrations perpendicular to plane of incidence.

4 0

2 years ago