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mezya [45]
3 years ago
7

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

4 s. What was the average acceleration during this run?
Physics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

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The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
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14. Which of the following is an example of
Murljashka [212]

Answer:

none of the above or screwdriver

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Two shotguns are identical in every respect except that one has twice the mass of the other. which gun will tend to recoil with
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The lighter one because it is lighter lol
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3 years ago
A carnival ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat
MaRussiya [10]

The acceleration of the person is 4.108 m/s^2.

What is acceleration?

The acceleration is the rate of change in the velocity in a unit of time.

Angular velocity: The change in angular displacement in a unit of time is called angular velocity.

Tangential velocity: The tangential velocity can be defined as the velocity of an object which is perpendicular to the radius in the rotational motion.

The relation between angular velocity ω, tangential velocity v, and radius r from the axis when the radius is perpendicular to the tangential velocity is,

v=ω*r

Given r=15 m, and ω=5 turns/minutes, substitute these values in the above formula.

Note: 1 turn/minute = 2π/60 rad/s.

v=(5 turns/minutes)*15 m

v=(5*2π/60 rad/s)* 15 m

v=7.85 m/s

Since the motion is circular, and the person is at the lowest point of the wheel, so the acceleration due to gravity will have no effect as it is perpendicular to tangential velocity here. The acceleration a for rotational motion is given by,

a=v^2/r

Substitute v=7.85 m/s, and r=15 m in this equation and solve it.

a=(2.094)^2/(15)

a=4.108 m/s^2

Learn more about acceleration here:

brainly.com/question/12550364

#SPJ4

5 0
2 years ago
The same mass, with a new string, is whirled in a vertical circle of the same radius about a fixed point. Find the magnitude of
Masteriza [31]

Answer:

256.3 N

Explanation:

Throughout the motion of the mass in the circular motion, the resultant of the Tension in the string and the weight of the body gives the net force (centripetal force) that is responsible for the acceleration of the mass in a circular motion.

At the top of the vertical circle of the motion of the mass, the tension is directed downwards towards the centre of the circle and the weight also is directed downwards towards the earth.

Net force = ma = T + W

ma = T + mg

But a = (v²/r) for circular motion,

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m = 2.31 kg

v = 7.77 m/s

r = 0.500 m

g = acceleration due to gravity = 9.8 m/s²

(2.31×7.77²)/0.5 = T + (2.31×9.8)

278.923 = T + 22.638

T = 278.923 - 22.638

T = 256.285 N = 256.3 N

Hope this Helps!!!

7 0
3 years ago
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