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tankabanditka [31]
3 years ago
10

Amit has found a rock in his backyard and would like to measure its mass, volume, and density. What tools can he use to perform

these measurements? meter stick, balance, and graduated cylinder graduated cylinder and balance meter stick and balance graduated cylinder and meter stick
Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer: a graduated cylinder and balance

A rock has an irregular shape. so, it would be difficult to measure using a meter stick. most convenient method to find the volume would be using a graduated cylinder. we can take a fixed volume of water in the cylinder. then drop rock into it. note the new raised volume. Subtract the initial volume from final to find the volume of rock.

Using balance, mass of the rock can be measured.

we can calculate the density using the following formula:

Density=\frac{Mass}{Volume}

Therefore, we need only two tools, a graduated cylinder and a balance to measure mass, volume and density of a rock.

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A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate
MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, m=1.16\times 10^{-26}\ kg

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

\dfrac{1}{2}mv^2=qV

v=\sqrt{\dfrac{2qV}{m}}

q is the charge on an electron

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}

v = 78608.58 m/s

v=7.86\times 10^4\ m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

qvB=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0
3 years ago
"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro
pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0
3 years ago
To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the
zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}

R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

8 0
3 years ago
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
3 years ago
The sun's core _____. A) is made of solid helium B) makes up most of the sun's size C) creates energy by fusing hydrogen into he
VashaNatasha [74]
I think the correct answer is C
6 0
3 years ago
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