If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

8 0

3 years ago

Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. It forms a pile in the shape of a right circu

Alona [7]

Answer:

dh/dt = 0.0566 feet / min

Explanation:

Volume per second, dV/dt = 30 cubic feet per minute

Diameter of cone = height of cone = h

at h = 15 feet, dh/dt = ?

Volume of cone is

V = πr²h

where, r be the radius of cone

According to question

radius of cone = half of diameter of cone = h /2

So,

$V=\pi \times \left ( \frac{h^{2}}{4} \right )\times h$

$V=\pi \times \frac{h^{3}}{4}$

Differentiate both sides with respect to h

$dV/dh = \frac{\pi }{4}\times 3h^{2}\times \frac{dh}{dt}$

by substituting the values

$30=\frac{3\times3.14}{4}\times 15\times 15\times \frac{dh}{dt}$

dh/dt = 0.0566 feet / min

7 0

4 years ago

If an object with more mass is pushed with the same force as an object with less mass the object with more mass will accelerate?

expeople1 [14]

Answer:

smaller acceleration, so lower change in velocity

Explanation:

To answer this question we examine the equation that relates mass with force and with acceleration: $F=m*a$ .

Since we want to know what happens to the acceleration, we solve for it in the equation: $a=\frac{F}{m}$

Notice that we are asked what happens when the force applied is the same, but now it is applied in an object with more mass (M).

We therefore would have to compare our initial form:

$a=\frac{F}{m}$ with the new one: $a=\frac{F}{M}$ wher the denominator is a larger quantity, therefore making our division/quotient smaller. Then, we conclude that the acceleration will be smaller, and therefore the change in velocity of the object will be lower.