WORTH 50 POINTSSSS!!!!!!!! don't lie either if you do I will report your answer and get my points back idc !!!!!!

Bromine is a liquid at room temperature. The volume of a sample of bromine is measured in a 50 beaker and a 100 ml beaker. How w

Stells [14]

Answer:

The comparisons are;

The height of the bromine in the 50 ml beaker will be twice that of the 100 ml beaker

The measurement of the volume with the 50 ml beaker will be more accurate than the measurement taken with the 100 ml beaker, because the differences in the height of the bromine in the 50 ml beaker is more obvious than the differences measured with the 100 ml beaker.

The actual volume of bromine in both beakers will be equivalent

Explanation:

The properties of a liquid are;

1) The volume of a liquid is relatively fixed at conditions that are suitable for it to remain in the liquid state compared to the volume occupied by a gas

2) A liquid will assume the shape of a container in which it is placed

3) The surface of a liquid in a container is flat due in order that the attractive forces between the molecules of the liquid at the surface and inside the body of the liquid should be in equilibrium

Therefore, given that the volume of the Bromine is measured in 50 ml beaker and a 100 ml beaker, there will be differences in the measured height of the same volume of bromine in each beaker.

5 0

2 years ago

What is the centripetal force that holds planets in orbit?

Fudgin [204]

<h2>Answer: Gravity force</h2>

If we approximate the orbit of the planets around the Sun to circular orbits with a uniform circular motion, where the velocity $\vec{V}$ is a vector, whose direction is perpendicular to the radius $r$ of the trajectory; the acceleration $\vec{a}$ is directed towards the center of the circumference (that's why it's called centripetal acceleration).

Now, according to Newton's 2nd law, the force $\vec{F}$ is directly proportional and in the same direction as the acceleration:

$\vec{F}=m.\vec{a}$

Therefore the net force resulting from the movement of a planet orbiting the Sun points towards the center of the circle, this is called Centripetal Force which is a central force that in this case is equal to the gravity force.

8 0

3 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

3 0

2 years ago

(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this sit

timurjin [86]

Answer:

1/2mv² = ke²

Explanation:

Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;

K.E = 1/2mass×velocity² i.e 1/2mv²

Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²

E.P = 1/2kx²

The equation describing the case where the kinetic energy is twice the elastic potential energy will be;

K.E = 2EP... 1)

Substituting the KE and EP formula into (1), we have;

1/2mv² = 2(1/2ke²)

1/2mv² = ke² which gives the required equation

8 0

3 years ago

A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall

Stells [14]

5,658 ft is your answer I believe

6 0

2 years ago