The amount of time required for 2 successive wave crests to pass a fixed point is called wave _______

Chintu and raven build a roccket, which moved from the earth to about 86m into the shy. It tales 3.7 seconds to reach the rocket

vaieri [72.5K]

Speed = (distance covered) / (time to cover the distance)..

From the ground to the highest point,
the rocket's AVERAGE speed is

(86 m) / (3.7 sec) = 23.2 m/s (rounded) .

5 0

2 years ago

The specific heat of iron is 0.45j/gk. The amount of heat needed to raise the temperature of the 12 g of iron by exactly 15k?

labwork [276]

Answer:

81 J.

Explanation:

From the question given above, the following data were obtained:

Specific heat capacity (C) = 0.45 J/gK.

Temperature change (ΔT) = 15 K

Mass = 12 g

Heat required (Q) =?

The heat required to raise the temperature of iron can be obtained as illustrated below:

Q = MCΔT

Q = 12 × 0.45 × 15

Q = 81 J

Therefore, the heat required to raise the temperature of the iron is 81 J.

8 0

3 years ago

How many centimeters are there in meter? b. 10 c. d 1000 e. 10000 100 2. A centimeter is equal to 1 inch b.½inch C 1/2.54 inch

astra-53 [7]

Answer:

a) There are 100 centimeters in 1 meter.

b) $\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

Explanation:

a) We have the conversion

1 m = 100 cm

So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

$1cm=\frac{1}{2.54}inch$

$\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

8 0

2 years ago

Help me and you will meet your favorite singer at your door.

Goshia [24]

Sunlight, because it provides a source of energy is the answer because plants also provide a food source (please put as brainliest answer)

4 0

2 years ago

Read 2 more answers

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.