Ask question

Ask question

All categories

4 years ago

8

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. suppose a 60 kg r

unner completes a 5.0 km race in 16 minutes.

1 answer:

professor190 [17]4 years ago

7 0

So it will take 5 minutes if a person running on a treadmill so it will be easy.

You might be interested in

Which term describes speed in a certain direction?

11Alexandr11 [23.1K]

Answer:

A. Velocity

Explanation:

Velocity is vector quantity thus has both magnitude and direction. It describes not only the speed but also the direction. Speed is scalar quantity so describes only speed but not direction. Energy has nothing to do with speed, acceleration describes change in velocity in a direction over time

8 0

2 years ago

Read 2 more answers

Dont use me for points please i need help, (circuits for physics)

zhannawk [14.2K]

Answer:

B, A, A, B

Explanation:

Just trust me on this one.

3 0

2 years ago

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

$d = ut + \frac{1}{2}at^2$

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

7 0

3 years ago

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m$

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0

3 years ago

which of the following objects is considered negatively charged A - one with excess electrons on the surface B - one with excess

Ne4ueva [31]

An object that could be considered as negatively charged would be when it has an excess of an electron in its atom. However, when it loses an electron, it could go back to its stable state which is "uncharged" or when there is an excess proton, it could be a positively charged object.

8 0

3 years ago