Calculate the speed of an object that moves 200 m in 4 s?

_______is stored energy

Lyrx [107]

Answer:

Kenetic

Explanation:

Potentiak is at the start

kenetic is stores energy being released

6 0

2 years ago

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What do repeated trails in an experiment allow scientists to do?

fredd [130]

Answer:

Explanation:

When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time. If we were testing a new fertilizer, we could test it on lots of individual plants at the same time.

3 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago

The force generated by a long muscle varies as it contracts through its range of movement. At which point is the greatest force

Sedaia [141]

Answer:

When the muscle is completely contract.

Explanation:

Remember the that maximum force of a muscle is when is completely contract. A characteristic of a muscle is that can contract and can relax in the opposite direction. In this way, when all the microfibers of the muscle are join together (they are contract) is when the maximum tissue force is applied.

With exercise the fibers of the muscles can grow or reproduce to strength the muscle.

8 0

3 years ago

Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r

bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}$

where,

T = Radiated surface temperature

$\lambda_{max}$ = peak wavelength

(a)

here,

$\lambda_{max}$ = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}$

(b)

here,

$\lambda_{max}$ = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}$

6 0

2 years ago