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Vlada [557]
4 years ago
12

A horizontal line on the time axis of a speed vs. time graph means an object is ___.

Physics
1 answer:
Svetllana [295]4 years ago
4 0
Choice-A is the correct one.
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The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases
Lady bird [3.3K]

Answer: 1.79\ rad/s^2

Explanation:

Given

Initial angular speed is \omega_1=26.2\ rad/s

Final angular speed is \omega_2=36.5\ rad/s

Time period t=5.75\ s

Magnitude of the fan's acceleration is given by

\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}

Insert the values

\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2

Thus, fan angular acceleration is 1.79\ rad/s^2

8 0
4 years ago
Read 2 more answers
What is the formula for displacement
pogonyaev
Displacement is usually given to you as it is, but you can also get displacement through velocity by Δd= Δv*t, where  <span>Δv is the change in velocity and t is the change in time. 

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4 0
3 years ago
What do we call the days when the sun is directly overhead the equator?
NikAS [45]

Answer:

Equinox

Explanation:

It is an astronomical phenomenon that is repeated twice a year, in summer and winter, and that is caused because the axis on which our planet rotates is slightly inclined with respect to the plane of translation, this is its trajectory around the Sun.

For this reason there are two equinoxes a year, a phenomenon completely opposite to the solstice, but also marks the beginning of the spring and autumn seasons.

The equinoxes occur around March 20 or 21 and September 22 or 23, and it is the days when the Sun is exactly over the equator, which makes day and night more or less the same duration. The word ‘equinox’ comes from Latin and means precisely ‘same night’. In those two moments the part of the Earth closest to the Sun is the equator.

8 0
3 years ago
The impulse given to a ball with mass of 2 kg is 16 N*s. If the ball starts from rest, what is its final velocity?
Alik [6]
The impulse is equal to the variation of momentum of the object:
I=\Delta p = m \Delta v
where m is the mass object and \Delta v = v_f - v_i is the variation of velocity of the object.

The ball starts from rest so its initial velocity is zero: v_i=0. So we can rewrite the formula as
I=m v_f
or 
v_f =  \frac{I}{m}

and since we know the impulse given to the ball (I=16 Ns) and its mass (m=2 kg), we can find the final velocity of the ball:
v_f =  \frac{16 Ns}{2 kg}=  8 m/s
7 0
3 years ago
Read 2 more answers
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
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