A sonar emits a sound signal of frequency 40000Hz, towards the bottom of the sea.

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

3 years ago

Igneous rocks are formed from _____.

Leni [432]

Answer:

magma

Explanation:

actually cooling and solidification of magma

7 0

3 years ago

Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms

kondaur [170]

Answer:

$\boxed {\boxed {\sf 100,000 \ Joules}}$

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

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

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

m= 2000 kg
s= 10 m/s

Substitute these values into the formula.

$E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

$E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)$

Multiply the first two numbers together.

$E_k= 1000 \ kg \times (100 \ m^2/s^2)$

Multiply again.

$E_k= 100,000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

$E_k= 100,000 \ J$

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 0

3 years ago

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

$g = \dfrac{GM}{r^2}$

mass of earth

$M = \dfrac{gr^2}{G}$

$M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}$

M = 5.972 x 10²⁴ kg

density = $\dfrac{mass}{volume}$

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378 x 10³)³

V = 1.08 x 10²¹ m³

density = $\dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}$

density = 5.52 x 10³ kg/m^3

density = 5520 kg/m^3

8 0

2 years ago

Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

so

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

2 years ago