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3 years ago

7

If you and your friend are riding your bikes side by side and your reference points are each other, are you in motion? Explain y

es or no

1 answer:

geniusboy [140]3 years ago

6 0

If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary

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A human hair is approximately 56 µm in diameter.

Ann [662]

Answer:

The diameter is 0.000056 m

Explanation:

Lets explain the relation between the meter and the micrometer

1 Meter is equal to 1000000 (one million) micrometers

1 micrometer = $\frac{1}{1000000}=\frac{1}{10^{6}}=10^{-6}$

The symbol of the meter is m

The symbol of micrometer is μm

A human hair is approximately 56 µm in diameter

We need to express this diameter in meter

To do that we divide this number by 1,000,000 or multiply it by $10^{-6}$

→ $\frac{56}{1000000}=0.000056$ 56 µm = 0.000056 m

→ OR

→ $56*10^{-6}=0.000056$

→ 56 µm = 0.000056 m

<em>The diameter is 0.000056 m</em>

4 0

3 years ago

At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0

3 years ago

Samanthawalks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the hor

Anna11 [10]

Answer:

Explanation:

Samantha walks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the horizontal part is for me what is the magnitude of displacement

Displacement is given by the straight line distance between P and Q. Displacement will be length of straight line joining P and Q

a semi circle with a radius of 2 m

Length of this straight line=4+diameter

=4+(2*2)

=8 m

7 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

2 years ago

Assemblage is ________________ a. a technique for casting where wax is carved then heated after a mold is made, then heated so t

labwork [276]

Assemblage is an additive process where artists construct work by putting together objects and attaches them in some way. The correct option among all the options that are given in the question is the second option or option "b". This kind of artistry actually gives a three dimensional impression to the work that is done by the artist.

7 0

3 years ago

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