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3 years ago

A snail slowly travels down the sidewalk at a pace of 4 cm/s. how long will it take the snail to reach a distance of 20 cm?

Physics

1 answer:

egoroff_w [7]3 years ago

8 0

It’s gonna take it 5 cm to reach the distance

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Identify the constants and variables in 2cm +4y

Harman [31]

constant = 2 and 4

variable = y

Explanation:

A constant is usually a symbol with a known value. In the expression, the constants are 2 and 4.

Their values are accurately known.

A variable is place holder and the value is not known. Variables are not static. They can take different numerical values as defined by an equation:

2cm +4y

the cm is the a unit

2 and 4 are constants

y is a variable that can assume any value

Learn more:

Experiment brainly.com/question/1621519

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3 years ago

If I was a scientist and I wanted to measure the intensity of an earthquake I would use

g100num [7]

Answer:

A seismograph

Explanation:

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2 years ago

48 grams 12cm^3, what would the density of the material be

Licemer1 [7]

Density is mass over volume:

$D = \frac{m}{V}$

In your case, mass is 48 grams and volume is 12cm^3

If you put that into the equation, you will get your density:

$D = \frac{m}{V} = \frac{48g}{12cm^{3} } =4g/ cm^{3}$

7 0

3 years ago

A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity

german

Answer: b

Explanation:

8 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago