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dimaraw [331]
3 years ago
13

A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do

es the puck have to travel in order to make it to the net?

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

\theta=90-40=50^o

From the triangle OAB

cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}

1340.57=35^2+25^2-x^2

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

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The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

  • m is mass of the rocket
  • v is the final velocity of the rocket
  • u is the initial velocity of the rocket
  • t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

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7 0
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How does the density of most metals compare to most non metals?
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Harder. Not compressible(unless using an extremely strong force). Non-metal have more of a chance of breaking than metals.
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When atoms are split, they release energy. This concept applies to (2 points)
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This applies to nuclear reactions, specifically nuclear fission.

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6 0
3 years ago
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan
pshichka [43]

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

6 0
3 years ago
A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0
OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

7 0
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