All categories

3 years ago

A sprinter ran a 100 meter dash in 12.5 seconds. What was his speed?'

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Speed is (pretty much) the same thing as velocity.

Velocity can be calculated by dividing the distance traveled over the time it took to travel said distance.

V=d/t

V=100 meters/12.5 seconds=8 meters per second

Velocity=8 m/s

You might be interested in

Which best describes the electric field created by a positive charge?

Vlad [161]

Its rays point away from the charge

3 0

3 years ago

Read 2 more answers

Pleaseeeeeeeeee helpppppppppppppppp

Tju [1.3M]

I feel like it could be A

8 0

2 years ago

The longest wavelength within the visible spectrum is _______ light.

Goryan [66]

<span>The longest wavelength within the visible spectrum is the red light. The answer is letter C. It is called visible light because it is the only light that can be seen by the human eye. Red light is the longest wavelength around 620 to 750 nanometer. It is followed by orange which has a wavelength of 590 t 620 nanometer. And then blue which has a wavelength of 450 to 495 nanometer. And the shortest wavelength is violet which has a wavelength of 380 to 459 nanometer. </span>

5 0

2 years ago

Read 2 more answers

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter $a=1.6m/sec^2$

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So $F=70\times 1.6=112 N$

So external force will be 112 N

6 0

3 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$

4 0

3 years ago