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allochka39001 [22]
3 years ago
14

A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the

cart? How fast will it be going at the end of the 4.5 seconds?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

t = time interval = 4.5 s

Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

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the amount of time for one particle of the medium to make one comlpete vibration cycle is a good description of
-Dominant- [34]
That's the definition of the PERIOD of the vibration.
It's exactly the reciprocal of the vibration's frequency.
7 0
3 years ago
Read 2 more answers
10. What is the acceleration of a 1000 kg car subject to a 500 N net force?
Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0
2 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?
Blababa [14]

The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

When you look at an object located 1 light year away from you, you see it as it was 1 year ago.

If a star located 10 light years away from us suddenly brightens, or dims, or explodes, we see the event <em>10 years later.</em>

7 0
3 years ago
Cables and conductors installed on the side of a pole or building shall be protected up to ___feet above finished grade.
Allisa [31]

Answer:

10 feet.

Explanation:

Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.

8 0
3 years ago
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