All categories

2 years ago

Write a hypothesis about the use of an object's physical

Physics

1 answer:

KengaRu [80]2 years ago

3 0

If the mass of the object and the volume of the object is determined;

Then, the density of the object is determined by taking the ratio of the mass and volume.

<h3>What is density of an object?</h3>

The density of an object is the ratio of the mass and volume of that object.

Mathematically;

Density = mass/volume

To determine the density of an object therefore, the physical characteristics of mass and the volume of the object are measured.

The mass of the object is obtained using a scale or a balance.

The volume of the object if a solid is obtained using a displacement bottle. If it is a liquid, a measuring cylinder is used.

The density of the object is then obtained by taking the ratio of the mass and the volume of the object.

In conclusion, the density of an object is determined from the volume and mass ratio.

Learn more about density at: brainly.com/question/1354972

#SPJ1

You might be interested in

An object is moving along a straight line at a

Dominik [7]

The answer for this question is 5 m

6 0

2 years ago

A ball of mass 6 kg is moving to the right with a velocity of 4 m/s when it strikes a 12 kg green block moving to the left at 5

Marat540 [252]

Answer:

Velocity of both masses after the collisio

Explanation:

Hope it will help

<h2><em><u>Brainlists please</u></em></h2>

3 0

2 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Mila [183]

Answer:

11.4 m/s

Explanation:

The expression for the Centripetal acceleration is :

$a=\frac{v^2}{R}$

Where, a is the accleration

v is the velocity around circumference of circle

R is radius of circle

In the given question,

a = g = Acceleration due to gravity as the car is at top = $9.81\ m/s^2$

v = ?

R = 13.2 m

So,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

8 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago

What happens to the free energy released as electrons are passed from photosystem II to photosystem I through a series of electr

lana [24]

<span>It is used to establish and maintain a proton gradient.</span>

6 0

3 years ago