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3 years ago

A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration

1 answer:

6 0

Average acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (beginning speed)

=  (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr

Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²

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A car is going through a dip in the road whose curvature approximates a circle of radius 200 m. At what velocity will the occupa

boyakko [2]

Answer:

Explanation:

The solution along with the diagram can be found in the attachment. Feel free to raise doubts.

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

A spring-loaded launcher has a mass of 0.60 kg and is placed on a platform 1.2m above the ground. The force of friction is negl

kobusy [5.1K]

Answer:

C The launcher will fall off the platform and land D/2 to the left of the platform because the launcher is twice the mass of the ball.

Explanation:

The figure is missing: you can find it in attachment.

We can apply the law of conservation of momentum to check that the launcher will leave the platform with a speed which is half the speed of the ball. In fact, the total initial momentum is zero:

$p_i = 0$

while the total final momentum is:

$p_f = m_l v_l + m_b v_b$

where

$m_l = 0.60 kg$ is the mass of the launcher

$m_b = 0.30 kg$ is the mass of the ball

$v_l$ is the velocity of the launcher

$v_b$ is the velocity of the ball

Since the total momentum must be conserved, $p_i=p_f$ , so

$0=m_l v_l + m_b v_b$

Therefore we find

$v_l = - \frac{m_b}{m_l}v_b = -\frac{0.30}{0.60}v_b = -\frac{v_b}{2}$

which means that the launcher leaves the platform with a velocity which is half that of the ball, and in the opposite direction (to the left).

Since the distance covered by both the ball and the launcher only depends on their horizontal velocity, this also means that the launcher will cover half the distance covered by the ball before reaching the ground: therefore, since the ball covers a distance of D, the launcher will cover a distance of D/2.