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3 years ago

What is the slope of the line?

Physics

1 answer:

coldgirl [10]3 years ago

6 0

(30, 5)
(10, 1)

change of y / change of x
= (30 - 10) / (5 - 1)
= 20 /4
= 5

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Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

3 years ago

A moving bumper car hits the back bumper of a stationary bumper car. The momentum of the stationary car increases. Which happens

kolbaska11 [484]

It decreases because it gave its momentum to the other car.

7 0

3 years ago

Read 2 more answers

A block rests on a ?at plate that executes vertical simple harmonic motion with a period of 0.58s.

SVETLANKA909090 [29]

Answer:

maximum amplitude = 0.08 m

Explanation:

Given that

Time period T= 0.58 s

acceleration of gravity g= 9.8 m/s²

We know that time period of simple harmonic motion given as

T = 2π/ω

0.58 = 2π/ω

ω = 10.83rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

9.8 = 10.83² A

A = 0.08m

maximum amplitude = 0.08 m

8 0

3 years ago

Read 2 more answers

A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as

frez [133]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

3 years ago