All categories

3 years ago

How long will it take a person walking at 2.1 m/s to travel 13 m?

Physics

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

This is a speed formula so you would use the formula speed=distance/time
You need to rearrange it to time=distance/speed
So you need to divide 13m by 2.1 m/s

You might be interested in

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Varvara68 [4.7K]

Answer:

Explanation:

A force of $55\text{ }N$ is acting on a wagon along the road. The wagon weights $7.5\text{ }kg$ . Acceleration of the wagon is given as $5.3\text{ }\frac{m}{s^{2}}$ .

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon $\times$ Acceleration of wagon

$F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N$

$F_{friction}$ has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0

3 years ago

According to the theory of plate tectonics, which forces cause the movement of plates in the earth's crust?

Sauron [17]

The correct answer is B.

4 0

3 years ago

Read 2 more answers

How long does for one rotation of earth

stellarik [79]

Answer: 24 hours

Explanation: One day is one rotation of the Earth, but it takes the earth 365 days to do one revolution or one year

8 0

3 years ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

5 0

3 years ago

A boy ties a stone to the end of a string which he then whirles above his head round a circular path of radius 15cm. if the ston

STALIN [3.7K]

Answer:

W = 12.568rads

V = 1.8852m/s

Explanation:

Data:

r = 15cm = 0.15m

t = 10s

Revolutions = 20

Frequency (F) = number revolutions/ time take complete it

F = 20 / 10 = 2Hz

Angular velocity (w) = 2 * 3.142 * F

Note: 3.142 = value of pi

W = 2*3.142*2 = 12.568 rads

Linear velocity (v) = w*r

V = 12.568*0.15 = 1.8852m/s

3 0

3 years ago