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viktelen [127]
4 years ago
12

If April launched a water balloon directly upwards with a speed of 40 m/s, how high would it be after 6 seconds? Use 10 m/s​

Physics
1 answer:
Volgvan4 years ago
3 0

Answer:

495m

Explanation:

ANSWER

u

stone

=10m/s (initial velocity of stone)

t=11s

∴H=−ut+

2

1

gt

2

(H=height of baloon)

H=−10×11+

2

1

×10×121=605−110=495m

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When a skateboarder rides down a hill, the normal force exerted on the skater by<br> the hill is
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Newton's second law allows us to find that the correct answer is:

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Newton's second law states that the net force and is propositional to the mass and acceleration of the body

For these problems it is essential to set a reference system, with respect to which to carry out the measurements, in this case we set a coordinate system with the x axis parallel to the plane and positive in the direction of movement and the y axis perpendicular to the plane.

In the attachment we can see a free-body diagram of the problem, let's work each axis separately

x-axis

         Wₓ -fr = m a

y-axis

         N - W_y = 0

         N = W_y

Where Wₓ and W_y are the components of the weight, fr the friction force that opposes the movement, m the mass and the acceleration of the body

let's use trigonometry to find the components of the weight

        cos θ = \frac{W_y}{W}

        sin θ = \frac{W_x}{W}

         W_y = W cos θ

         Wₓ = W sin θ

we substitute

          N = mg cos θ

From this equation we can see that the normal is less than the weight of the body.

In conclusion using Newton's second law we find that the correct answer is:

  • Less than th weight of the skater

 

Learn more about Newton's second law here:

brainly.com/question/13685393

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A 90kg person pushes a 25kg cart 15m up a ramp using a force of 80N. The top of the ramp is 3m above ground. It takes 8 seconds
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convert 1 milligram to 1 gram:

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so... 2 milligram is 0.002

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A skydiver jumps out of a hovering helicopter, so there is no forward velocity. Ignore wind resistance for this exercise. 1. Wha
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Initially, like any falling object, a skydiver's downward acceleration is 9.8 meters/seconds^2, or about 28-35 feet per second squared. This acceleration reduces over a few seconds and approaches zero as the skydiver reaches terminal velocity.
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Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0
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Answer:

Ex=  -17.1 N/C

Ey =  +26.9 N/C

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q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

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Calculation of the field at point P due to the load q₂

d_{2} = \sqrt{1.30^{2}+0.450^{2}  }

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Ex=E₂x= -17.1 N/C

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