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2 years ago

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A ruined in a race decides to accelerate right up to the moment he crosses the line He is the initially travelling at 5m/s and a

ccelerates at 0.4m/s² for 5 sec. find
a) the distance he covers
b) his final velocity

1 answer:

docker41 [41]2 years ago

6 0

Answer:

S = V t + 1/2 a t^2 = 5 m/s * 5 s + .2 m/s^2 * 25 s^2 =

25 m + 5 m = 30 m distance traveled

Vf = V + a t = 5 m/s + .4 m/s^2 * 5 s = (5 + 2) m/s = 7 m/s final velocity

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Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

Answer: Option A

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According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

$Force = Mass \times Acceleration$

So,

$Acceleration=\frac{\text {Force}}{\text {Mass}}$

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

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3 years ago

Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?

Naya [18.7K]

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

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3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

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Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

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Part c)

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Part d)

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Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

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$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

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Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

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Part e)

Angle of projection is given as

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$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

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$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

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dv = a_avg dt

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