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3 years ago

A car leaves the pit after a refueling stop and accelerates uniformly to a speed of 62.2 m/s in 5.0 s to rejoin the race.

Physics

1 answer:

MakcuM [25]3 years ago

5 0

Answer:

The car's acceleration is: a = 12.44 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} = v_{o}+a*t$

where:

Vf = final velocity = 62.2 [m/s]

Vo = initial velocity = 0 (because the car stars from the rest)

a = acceleration [m/s²]

t = time = 5 [s]

62.2 = 0 + a*5

a = 12.44 [m/s²]

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A wave completes one vibration as it moves a distance of 2 meters at a speed of 20 meters per second. What is the frequency of t

Gemiola [76]

Answer:

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Explanation:

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6 0

3 years ago

Assume that the operating cost of a certain truck (excluding driver's wages) is 12+x/6 cents per mile when the truck travels at

podryga [215]

Answer:

$x = 60 mph$

Explanation:

Given that the operating cost is

$c = 12 + \frac{x}{6}$ cents per mile

total miles covered is given as

$d = 400 miles$

so total cost of drive is given as

$C = (12 + \frac{x}{6})(4)$ $

time taken by the truck to move the distance is given as

$t = \frac{400}{x}$

So total earnings of the driver is given as

$E = \frac{400}{x} \times 6$ $

now total profit of the driver is given as

$P = \frac{2400}{x} - (48 + \frac{2x}{3})$ $

to maximize the profit we have

$\frac{dP}{dx} = 0$

$-\frac{2400}{x^2} + \frac{2}{3} = 0$

so we have

$x = 60 mph$

8 0

3 years ago

A motorcycle has a mass of 250kg. It goes around a 13.7 m radius turn at 96.5 km/hr. What is the centripetal force?

topjm [15]

Answer: c) 1.31 × 10⁴N

Explanation:

Centripetal force is a force that causes a body to move in a circular path. The body possesses a centripetal acceleration.

According to newtons first law

Force = mass (m) ×acceleration (a)

Acceleration of a body moving in a circular path = v²/r where;

v is the velocity and r is the radius

Force = MV²/r

Given m = 250kg v = 96.5km/hr r = 13.7m

We need to convert 96.5km/hr to m/s

= 96.5km ×1000/1hr × 3600

= 96,500/3,600

v = 26.8m/s

Force = 250 × 26.8²/13.7

Force = 13,106.5N

The centripetal force = 13,106.5N

= 1.31 × 10⁴N

4 0

3 years ago

A cart is pulled horizontally with a 156.6 N force to the right at a 67 degree angle with respect to the ground. The cart is mov

ella [17]

Answer:

b) Vertical= 144.15N [up]

horizontal= 61.19N [right]

c) -61.19N [Left]

D) 9.8m - 144.15 = Fn

Explanation:

a) to draw the free body diagram, you would draw a square to represent the object, being the cart. The forces acting on it are Force of gravity (Fg going down), Normal force (Fn going up), Force applied (Fa going up right), and Force of friction (Ff,going to the left, opposite of Fa).

b) So the applied force is 156.6N, and has an angle of 67. the 156.6N is the hypotenuse, so you would need to find the opposite side (vertical component), and adjacent side (horizontal component):

vertical:

sinθ = opp / hyp

sin67 = opp / 156.6

opp = 144.15N [up]

horizontal:

cosθ = adj / hyp

cos67 = adj / 156.6

adj = 61.19N [right]

c)Since the cart has constant velocity, the acceleration is 0, meaning no net force (fnet = ma, a = 0, fnet = 0), so the horizontal component of the applied force is equal to the frictional force. Heres why mathematically. (X DIRECTION ) :

Fnet = Fa + Ff

0 = Fa + Ff

-Ff = Fa.

So the force of friction has the same magnitude, but opposite direction. So since the horizontal component of applied force was 61.19 N, the frictional force will be -61.19N

d) So now we are looking at the Fnet in the y direction, which is also 0 because the car is remaining on the ground, its not going up nor down:

Fnety = 0

Fnet is composed of Fg, Fn and Fay (force applied in y direction);

fnet = Fn + Fay + Fg. Fn and Fay are both up, and Fg is down, so u need to subtract Fg:

fnet = Fn + Fay - Fg.

0 = Fn + 144.15 - mg

mg - 144.15 = Fn

9.8m - 144.15 = Fn

Since it didnt give you a mass thats how much you could simplify it down to, in order to get the Fn

6 0

3 years ago

Subtracting a negative number follows the same process as a positive number

Sav [38]

This would be true!

When you subtract a positive number, it goes to the left of the number line. When you subtract a negative number, it goes to the left of the number line.

The thing that confuses most people is that the first number (when negative) continues to increase as a negative. It follows the same process, just different signs.

I hope this helps!
~cupcake

6 0

3 years ago