Which best describes most covalent compounds? O soft O brittle 3 O cold O warm

A singly charged positive ion that has a mass of 6.07 ? 10-27 kg moves clockwise with a speed of 1.18 ? 104 m/s. the positively

mezya [45]

B = 0.018 T Ans,

Since, it is moving in a circular path, thus, centripetal force will act on it i.e.
F = $\frac{ mv^{2} }{r}$
where, m is the mass of the object, v is the velocity and r is the radius of circular path.

And, since a positive charge is moving, it will create magnetic force which is equal to F = qvB
where q is the charge, v is the velocity of the particle and B is the magnetic field.
Now, the two forces will be equal,
i.e. $\frac{ mv^{2} }{r}$ = qvB
⇒ $\frac{ mv }{r}$ = qB
⇒B = $\frac{ mv }{qr}$
putting the values, we get,

use q = 1.6 * 10^ -19
⇒ B = 0.018 T

8 0

2 years ago

A quantity of gasoline injected into car engine has a potential energy of 100J. All of this energy is not converted into useable

neonofarm [45]

Answer:because some energy are lost due to friction

Explanation:some energy are lost due to friction

5 0

2 years ago

5. It is the year 2085, and the world population has grown at an alarming rate. As a space explorer, you have been sent on a ter

shtirl [24]

Answer:

Answer:

It has sunlight, oxygen, water, and fertile soil.

Explanation:

Sunlight: provides plants (producers) with energy so that the ecosystem is sustainable. Also provides heat and warmth so the planet does not freeze.

Oxygen: obvious. required for human survival and survival of most carbon-based organisms.

Water: required for plant growth and sustainability of carbon-based life.

Fertile soil: a place where plants can grow and produce energy.

I will not write an entire description because your teacher may accuse you of plagiarism and use this topic as proof.

Explanation:

3 0

2 years ago

Read 2 more answers

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

2.78m/s²

Explanation:

Complete question:

A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

Hence the acceleration of the box as it slides down the incline is 2.78m/s²

5 0

2 years ago