Please choose all that describe a series circuit.

When we kept the Earth's mass the same, but shrank its size, we saw that had an effect on its escape speed. Albert Einstein used

nadezda [96]

Answer:

Check the explanation

Explanation:

The escape velocity is the velocity needed by any object to overcome the gravitational force of the planet on which it’s present. Now we know that the gravitational force is directly proportional to the mass of the planet and inversely proportional to the distance of the object from the center of planet.

If we keep the mass of earth constant and decrease the size of the earth than this will decrease the distance between the object and the center of the earth and thus the gravitational force that will act on the body will increase substantially which will in turn increase the value of the escape velocity.

The value of escape velocity will keep on increasing as the size of the earth will shrink till it reaches to a point when the value of escape velocity becomes more than the speed of light and since it’s impossible to travel with a speed greater than the speed of light and therefore at this point it will become impossible for a spacecraft to escape the earth.

7 0

3 years ago

Whta is the mechanical advantage when the effort force is 20 and the resistance force is 5

pogonyaev

The mechanical advantage would be 4. 20/5=4

7 0

3 years ago

the book has a mass of 2.5 kg. What net force must act on the book to mak it accelerate to the left at a rate of 7.0m/s2?

Reil [10]

Answer:

17.5 N

Explanation:

<h2>Given :</h2>

Mass (m) = 2.5 kg
Acceleration (a) = 7.0 m/s²

<h2>To calculate :</h2>

Force exerted (F)

<h2>Calculation :</h2>

→ F = (2.5 × 7.0) N

→ F = 25/10 × 7 N

→ F = 5/2 × 7 N

→ F = (5 × 7)/2 N

→ F = 35/2 N

→ <u>F</u><u> </u><u>=</u><u> </u><u>1</u><u>7</u><u>.</u><u>5</u><u> </u><u>N</u><u> </u><u>towards</u><u> </u><u>left</u>

Hence, 17.5 N of net force must act on the book to make it accelerate to the left.

5 0

3 years ago

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$ ,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0

3 years ago

A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of

Mumz [18]

A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight).

F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N

a = (735.75 N)/(500kg)= 1.4715 m/s^2

4 0

3 years ago