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nikdorinn [45]
2 years ago
10

The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point

should be used to find out how far Ned traveled when he went from the pet store to the swimming pool?
Physics
2 answers:
aleksley [76]2 years ago
6 0

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

vodomira [7]2 years ago
6 0

Answer:

i think its w

Explanation:

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A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

5 0
2 years ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the
Blizzard [7]
<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

6 0
2 years ago
The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo
agasfer [191]

As we know by Doppler's effect of light we have

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

3.01 \times 10^{-3} = \frac{v}{c}

here we have

v = 904.3 km/s

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

3 0
2 years ago
A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the
Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

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Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the  microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure P_r = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

P_r = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

or

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

or

F = 2 × 10⁻³ N

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