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2 years ago

Describe a situation in which different units of measure could cause confusion.

1 answer:

6 0

When an experiment happens in the USA and the test of the world read it conversions may go wrk v and think that the experiment is false

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A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract

Ivan

Answer:

Explanation:

Let v be the linear velocity , ω be the angular velocity and I be the moment of inertia of the the puck.

Kinetic energy ( linear ) = 1/2 mv²

Rotational kinetic energy = 1/2 I ω²

I = 1/2 m r² ( m and r be the mass and radius of the puck )

Rotational kinetic energy = 1/2 x1/2 m r² ω²

= 1/4 m v² ( v = r ω )

Total energy

= Kinetic energy ( linear ) + Rotational kinetic energy

= 1/2 mv² + 1/4 m v²

= 3/4 mv²

rotational K E / Total K E = 1/4 m v² / 3/4 mv²

= 1 /3

So 1 /3 rd of total energy is rotational K E.

3 0

2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

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7 0

2 years ago

an iron rod of length 100m at 10 degree Celsius is used to measure a distance of 2km on a day when the temperature is 40 degree

german

Answer:

0.68 m

Explanation:

α = dL / L1*(dT)

dL = L1(dT) * α

Initial length, L1 = 100

Chang in Length =dL

α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length

α of iron rod = 1.13 * 10^-5 k

dL = 100(40 - 10) * 1.13 * 10^-5

dL = 100(30) * 1.13 * 10^-5

dL = 3000 * 1.13 * 10^-5

dL = 3390 * 10^-5

dL = 0.0339 m

Error :

Distance measured = 2km = (2 * 1000) = 2000m

[Distance measured / (initial length + change in length)] × change in length

Error = (2000 / (100 + 0.0339)) * 0.0339

Error = (2000 / 100.0339) * 0.0339

Error = 19.993222 * 0.0339

Error = 0.6777702

Error = 0.68 m

4 0

2 years ago

Determine the values of m and n when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,00

yuradex [85]

Explanation:

Mass of the Earth is equal to,

$m=5,970,000,000,000,000,000,000,000\ kg$

Any number can be written in the form of scientific notation as :

$N=m\times 10^n$

m is the real number

n is any integer

Mass of the earth can be written in the form of scientific notation as :

$m=5.97\times 10^{24}\ m$

Here,

m = 5.97

n = 24

Hence, this is the required solution.

7 0

2 years ago

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to

jeyben [28]

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

V = i (r + R)

We replace we calculate

r + R = V / i

R = v / i - r

R = 10/12 -0.04

R = 0.793 Ω

We calculate the power supplied

P = V i = I² R

P = 12² 0.793

P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

P = E / t

t = E / P

t = 100 10³/114

t = 877 s

The time to boil the water is 877 s

4 0

2 years ago