Ask question

Ask question

All categories

3 years ago

6

You are holding a rock tied to a string, which is an example of a pendulum. Identify and explain the proper term for each part o

f the pendulum in this example.

1 answer:

marshall27 [118]3 years ago

3 0

A pocket watch would be an example

You might be interested in

Damon purchased a pair of sunglasses that were advertised as being polarized. Describe how Damon could test the sunglasses to ve

Andre45 [30]

Answer:

To verify that they're polarized, he could hold the two lenses perpendicular (90 degrees) to each other, one lens in front of the other, and point it at a light source. If no light passes through then the lenses are polarized

6 0

4 years ago

A 20 liter cylinder of helium at a pressure of 150 atm and a temperature of 27ÁC is used to fill a balloon at 1.00 atm and 37ÁC.

djyliett [7]

<u>Answer</u>

D) 3100 Liters

<u>Explanation</u>

To get the volume if the balloon you need to use the combined equation of the low of gases.

P₁V₁/T₁ = P₂V₂/T₂

(20×150)/(27+273) = (1×V₂)/(37+273)

3000/300 = V₂/310

10 = V₂/310

V₂ = 10 × 310

= 3100 Liters

7 0

3 years ago

Read 2 more answers

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

sp2606 [1]

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$

$T_{min}=1.805*10^{-6}min$

$T_{hours}=3.0083*10^{-8}hours$

From the Question we are told that

Velocity $v=13.0m/s$

Distance $d=120 km$

Generally the equation for the Time is mathematically given as

$t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s$

Therefore

$T_{min}=1.083*10^{-4}s/60$

$T_{min}=1.805*10^{-6}min$

And

$T_{hours}=T_{min}/60$

$T_{hours}=3.0083*10^{-8}hours$

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

4 0

3 years ago

A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little l

Y_Kistochka [10]

The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters

4 0

4 years ago

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago