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Blizzard [7]
3 years ago
5

Anybody wanna help? (Picture Included?)

Physics
2 answers:
mash [69]3 years ago
5 0

Answer:

Corn kernels-popcorn

Snow falling-ice

Water-rain falling

Lava-volcanic rock

Bread-toast

Sand-glass

answers

Explanation:

Sindrei [870]3 years ago
3 0
Corn kernels-popcorn
Snow falling-ice
Water-rain falling
Lava-volcanic rock
Bread-toast
Sand-glass
answers
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Two blocks are connected as shown in the diagram below. Assume that the ramp is frictionless. Draw the force diagram for the blo
cluponka [151]

Answer:

diagram: see image, x-component: 84.3 N, acceleration: 4.38 m/s^2

Explanation:

(see image for further explanation)

5 0
3 years ago
If the radius of the atom is the distance from point A to D, where is the MOST likely location of the LEAST concentration of mas
lana [24]

Answer: B

Explanation:

3 0
4 years ago
A ball rolls from x = - 5m to x = 0m in 1 second . What was its average velocity? (Units = m/s) Don't forget: velocities and dis
katrin2010 [14]

Answer:

+ 5 m/s

Explanation:

change in displacement = ΔX=final position - initial position

                                            ΔX = 0-(-5) =0+5 =+ 5 m

average velocity =  ΔX/t

                             = +5/1

                             = + 5 m/s

positive sign shows that ball rolls towards right

8 0
3 years ago
A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a
Zolol [24]

Answer:

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta} (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L

For small oscillations, the equation can be re-arranged into the following form:

E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L

Where:

\theta = \frac{A}{L^{2}}, measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is 4\theta and the total energy of the pendulum is:

E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L

The factor of change is:

\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}

3 0
3 years ago
The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal
kogti [31]

Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

3 0
2 years ago
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