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3 years ago

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Suppose a rock is hoisted up with a rope. The rock experiences 10 N of tension force upward and 10 N of gravitational force down

ward. The total net force on the rock is 20 N. True False

2 answers:

nika2105 [10]3 years ago

5 0

False. Since the forces are pulling in equal and opposite directions, the net force is 0.

LiRa [457]3 years ago

5 0

10N of tension upwards is <em>positive</em>

10N of gravitational Force downwards is <em>negative</em>

Net Force = The sum of all forces

$F_{net} = F_{tension} + F_{gravity}$

$F_{net} = 10N + (-10N) = 0$

So the answer is false.

Extra Info: If the net force was 20N there would be a tensional force of 30N moving the rock upwards.

30N - 10N = 20N

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Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

$v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\$

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

$v^{2}=u^{2}-2a(y-2.2)\\$

if we substitute values we have

$v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m$

c. the time it takes to arrive at 1.83m is obtain by using the equation below

$1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37$

if we insert the values, we solve for t , hence t=1.43secs

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3 years ago

The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a sp

AlekseyPX

Answer:

$t = 0.93 s$

Part b)

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Part c)

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Explanation:

The two components of the velocity of the ball is given as

$v_x = 4.9 cos29$

$v_x = 4.28 m/s$

$v_y = 4.9 sin29$

$v_y = 2.37 m/s$

Part a)

now we know that the displacement in y direction is given as

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