The phenomenon of water sticking to a surface, such as a window pane or

If it takes Tanya 9.4 hours to get to her cousin's house while driving an average of 40 mph, how far away does her

Rufina [12.5K]

Answer:

376 miles

Explanation:

speed = distance / time

therefore distance = speed * time

distance = 40 miles/hr * 9.4 hr

then

distance = 375 miles

5 0

3 years ago

An automobile with a mass of 1180 kg is traveling at a speed v = [06] ____________________ m/s. a. What is its kinetic energy in

iogann1982 [59]

Complete Question:

An automobile with a mass of 1180 kg is traveling at a speed v =2.51 m/s. What is its kinetic energy in SI units? What speed (m/s) must an 82.7-kg person move to have the same kinetic energy? At what speed (m/s) must is 12.1-g bullet move to have the same kinetic energy? What would be the speed (m/s) of the automobile if its kinetic energy were doubled?

Answer:

a) 3717.1 J b) 9.48 m/s c) 783.8 m/s d) 3.55 m/s

Explanation:

a)

By definition, the kinetic energy of a mass m with a speed v, is as follows:

$K = \frac{1}{2} * m *v^{2}$

if m= 1180 Kg, and v= 2.51 m/s, the kinetic energy can be calculated as follows:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*(2.51 m/s)^{2} = 3717.1 J$

b)

If the kinetic energy must be the same, and m= 82,7 Kg, we can write the following expression:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 82.7 kg*((v)(m/s))^{2} = 3717.1 J$

We can solve the above equation as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{82.7kg} } = 9.48 m/s$

c)

If K remains the same, and m = 12.1 g = 0.0121 kg (in SI units). we can solve for v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{0.0121kg} } = 783.8 m/s$

d)

Now, if the kinetic energy were doubled, we would have the following equation:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*((v) m/s)^{2} = 3717.1 J * 2 = 7434.2 J$

We can solve for the new speed v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*7434.2J}{1180kg} } = 3.55 m/s$

8 0

3 years ago

If a simple machine reduces the strength of a force, what must be increased?. . A. the distance that the force move. . B. the wo

Karolina [17]

"The distance that the force moves" is the one among the following choices given in the question that must be increased, if <span>a simple machine reduces the strength of a force. The correct option among all the options that are given in the question is the first option or option "A". I hope the answer helped you.</span>

3 0

3 years ago

Read 2 more answers

the table below shows the effect conditions on a car stopping distance. in witch condition was the car travelling most slowly

nikdorinn [45]

Answer: Car 5

Explanation:

Higher speeds mean higher braking times so the car with the highest braking distance was travelling the the fastest and the car with the lowest braking distance was traveling the slowest.

Braking distance = Stopping distance - Thinking distance.

Car Braking distance

1 38

2 50 = 65 - 15

3 75

4 75

5 14 = 23 - 9

<em>Car 5 had the lowest braking distance so was going the slowest. </em>

7 0

3 years ago

A 50.0 kg object is moving at 18.2 m/s when a 200 N force

gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a = force ÷ mass

a = $\frac{-200}{50}$ = -4 m/s²

we get here now required time that is

required time = $\frac{V_{(final)} - V_{(initial)}}{a}$ ...............1

put here value

required time = $\frac{12.6-18.2}{-4}$

so distance will be

distance = $\frac{V_{(final)}^2 - V_{(initial)}^2}{2a}$ ........2

distance = $\frac{12.6}^2 -{18.2}^2}{2\times (-4)}$

distance = 21.56 m

7 0

3 years ago