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Ivahew [28]
3 years ago
6

Which equation can be used to solve for acceleration? ​

Physics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

the answer is the first one.

Explanation:

one way to find out is if you look at the other equations they are mixed matched. For an example a=d/t should be s= d/t.

antoniya [11.8K]3 years ago
3 0
A = d/t
hope this helps x
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When using the magnification equation, a value greater than 1 as the solution for M indicates that the image ? is larger than th
Nadusha1986 [10]
M= Height of image/height of the object

If, M>1, then height of image>height of object.

And therefore, image is larger than the object.
7 0
3 years ago
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Mary takes 6.0 seconds to run up a flight of stairs that is 102 meters long. If Mary's weight is 87 newtons, what power has Mary
fredd [130]

Answer: 1479watts

Explanation:

Power is defined as the energy expended or work done in a specific time.

Mathematically,

Power = Workdone/time taken

Since work done is force × distance

Power = force × distance/time

Force = Mary's weight = 87N

Distance = height of the flight = 102meters

Time = 6.0seconds

Substituting in the formula we have;

Power = 87 × 102/6

Power = 1,479watts

Note that the time must be in seconds before usage. If its given in minutes, you will have to convert to seconds

8 0
3 years ago
The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
3 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
2 years ago
A car is traveling down the highway at speed 5 m/s when the driver slams on the brakes and skids to a stop in a distance 80 m. A
Murljashka [212]
The car is very incredibly fast
8 0
3 years ago
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