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xenn [34]
3 years ago
5

What is the relationship between speed and the slopes of the lines on a graph

Physics
1 answer:
Strike441 [17]3 years ago
8 0
There relationship can be the same in a way because both of those slopes can increase
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A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the char
Kamila [148]

Answer:

B. 3.1 × 10^5 V

Explanation:

8 0
3 years ago
Read 2 more answers
SOMEONE PLEASE HELP!!
Arisa [49]

Hello!

\large\boxed{\text{C. 7,350,000 J}}

Use the equation:

PE = mgh

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0
3 years ago
Ohm's law states which relationship between electrical quantities? a. Volts = current x resistance b. Volts = current divided by
miv72 [106K]

Answer:

a. Volts = current x resistance

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Hence, Ohm's law gives the relationship between voltage, current and resistance of an electric circuit.

7 0
3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
a 60g tennis ball travelling at 30m/s hi a wall and bounce back at 20m/s.calculate (a)momentum of the ball before impact (b)mome
frez [133]

Explanation:

the formula for momentum is denoted by p=mv where p is momentum, m is mass and v is velocity. thus, the velocity before impact would be 0.060 x 30 = 1.8 kg/ms

the second one would just be 0.060 x 20 0.72kg/ms

I'm not 100 percent sure this is correct but yeah

6 0
3 years ago
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