3 years ago

If you run at 8m/s for 15 minutes how far will you go

Physics

2 answers:

enyata [817]3 years ago

8 0

Answer:

<h2>hope it helps you see the attachment for further information ......✌✌✌✌✌</h2>

sashaice [31]3 years ago

6 0

Answer: 7200 m

Explanation: The solution is, first convert 15 minutes to seconds.

15 mins x 60 s / 1 min = 900 s

Use the formula for speed which is v= d/t then derive for d.

d = vt

= 8 m/s ( 900s)

= 7200 m

You might be interested in

What causes the change from day to night?

stepladder [879]

Earths Rotation
Earth's rotation is the rotation of Planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north<span> pole star Polaris, Earth turns counter</span>clockwise<span>. </span>
"AB84"

7 0

3 years ago

Read 2 more answers

An astronaut weighs 104 newtons on the moon, where the strength of gravity is 1.6N/kg. What is her mass?

Leona [35]

Her mass is 65 kilograms

5 0

3 years ago

Read 2 more answers

Do you think a moving skateboard has energy? Why or Why not? What about a skateboard that is not moving?

puteri [66]

Answer:

Yes

Explanation:

Because if you push it, the skateboard has kinetic/potential energy.

3 0

3 years ago

Read 2 more answers

You step onto a hot beach with your bare feet. A generated in your foot, travels through your nervous system at an average speed

madreJ [45]

Answer: 12.44 millisecond

Explanation:

Based on the information given, to know the time taken for the impulse, which travels a distance of 1.58m to reach the brain will be:

Time taken = Distance/Speed

Time taken = 1.58/127

Time taken = 0.01244

Time taken = 12.44 millisecond.

7 0

3 years ago

Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,

Elza [17]

The electric field is zero at x = -16.45cm

Data;

q1 = 3.4 μC
q2 = -2.0 μC
distance = 5cm

<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

$k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\$

Let's solve for x using mathematical methods.

$\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0$

Solving the above quadratic equation;

$x = -16.45cm$

The electric field is zero at x = -16.45cm

Learn more on electric field at a point here;

brainly.com/question/1592046

brainly.com/question/14372859

8 0

3 years ago