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3 years ago

7

Betelgeuse is how many times larger than the sun

1 answer:

Aleksandr-060686 [28]3 years ago

3 0

Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).

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Which of the following examples illustrates static friction?

vivado [14]

Answer:

A box sits stationary on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0

3 years ago

An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat

KATRIN_1 [288]

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

8 0

3 years ago

Read 2 more answers

In what circuit does an increase in resistance mean an increase in current?

skad [1K]

I think in parallel circuits.

6 0

2 years ago

The first version of a product, which is built and tested during the third stage of technological design, is called a

Dimas [21]

Answer:

The third stage of parturition is called " after-birth".

Explanation:

8 0

2 years ago

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To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$u_{x}=u\cos\theta$

Put the value into the formula

$u_{x}=24\cos(31)$

$u_{x}=20.5\ m/s$

We need to calculate the vertical velocity

Using formula of vertical velocity

$u_{y}=u\sin\theta$

Put the value into the formula

$u_{y}=24\sin(31)$

$u_{y}=12.3\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{u_{x}}$

Put the value into the formula

$t=\dfrac{44}{20.5}$

$t=2.1\ sec$

We need to calculate the vertical height

Using equation of motion

$h=u_{y}t+\dfrac{1}{2}at^2$

Put the value into the formula

$h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2$

$h=4.2\ m$

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

$d=h-3.05$

Put the value of h

$d=4.2-3.05$

$d=1.15\ m$

Hence, The distance by the ball clear the crossbar is 1.15 m

7 0

3 years ago