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2 years ago

If a star is found directly above the sun on the h-r diagram, what can you conclude about its size?.

Physics

1 answer:

hammer [34]2 years ago

7 0

Answer:

It is larger than the Sun.

Explanation:

Brainliest pls

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A sound wave has a frequency of 615 hz in air and a wavelength of 0.54 m. what is the temperature of the air? assume the velocit

Leno4ka [110]

Answer is: T = -4.706°C

Explanation: Please see the image attached with this answer.

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2 years ago

What is the name of a very porous igneous rock that is so light that it floats?

Sholpan [36]

The answer is pumice rock

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2 years ago

32.What is the area under the curve between 0 seconds and 2 seconds?

Mademuasel [1]

Explanation:

32. The area under the curve from t = 0 and t = 2 is the area of the triangle. The base of the triangle is 2 seconds, and the height of the triangle is 2 m/s. The area is therefore:

A = ½ bh

A = ½ (2 s) (2 m/s)

A = 2 m

33. The car is not moving when the velocity is 0 m/s. This occurs at t = 0, t = 6, and t = 10.

3 0

3 years ago

A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance

djyliett [7]

Answer:

you can learn from here

https://www.toppr.com/ask/en-bd/question/a-car-is-moving-with-a-velocity-of-10-ms-the-driver-sees-a-wall/

6 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago