The speed at which a wave is transmitted depends on____

How much energy or stopping power is needed to bring a car to a stop from 100 mph?

Fed [463]

I think 100 mph pushing the car the opposite direction

3 0

3 years ago

The graph shows the number of beans eaten by a

il63 [147K]

Answer:

You didn't show a graph

Explanation:

6 0

4 years ago

Read 2 more answers

A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the

koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of proton = 1.67 x 10⁻²⁷ kg
v is speed of the proton = ?

<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

Learn more about momentum here: brainly.com/question/7538238

#SPJ4

4 0

2 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.

Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0

4 years ago

The speed at which a wave is transmitted depends on_____