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lina2011 [118]
3 years ago
6

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
Physics
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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Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it
stich3 [128]

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

6 0
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)

d = 3.98 m

Part c)

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v_y = -11.49 m/s

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

\Delta y = 6.4 m

so we have

\Delta y = v_y t + \frac{1}{2}gt^2

6.4 = 2.37 t + 4.9 t^2

t = 0.93 s

Part b)

Distance of the ball in x direction of the motion is given as

d = v_x t

d = 4.28 \times 0.93

d = 3.98 m

Part c)

In x direction the velocity will remain the same always

v_x = 4.28 m/s

while in Y direction we can use kinematics

v_y = v_{oy} + at

v_y = -2.37 - 9.81(0.93)

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Answer:

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3 years ago
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