Describe an experiment that can be used to determine the resistance of a metallic conductor

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

What's the difference between electromagnetic waves and electromagnetic spectrum?

White raven [17]

The difference between the two is, well for one

Spectrum: The entire range that the "waves" could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

It may confuse you but it makes sense to me (Sorry)

4 0

3 years ago

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

mylen [45]

The distance covered by the acorn is 3.136 m.

Explanation:

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

$s=ut+\frac{1}{2}gt^{2}$

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

$s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}$

So the distance covered by the acorn is 3.136 m.

8 0

3 years ago

Please help me this is timed . Find x if a = 3.0 m/s^2

storchak [24]

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

5 0

3 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

3 years ago