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3 years ago

9

You may have noticed that there was a big difference in the results of

1 answer:

vodomira [7]3 years ago

8 0

Answer:

For conduction you don't need position. For convection you do.

Explanation:

you just needed to look it up LOL <3

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Sound is faster than light that is why we hear thunder before we see lightning

Umnica [9.8K]

Answer:

false

Explanation:

sound travels slower than light. that is why we see lightning before we hear the thunder

8 0

3 years ago

A motorcycle starting from rest covers 200 metre distance in 6 second. Calculate final velocity and acceleration of the motorcyc

QveST [7]

Answer:

Explanation:I don't say you must have to mark my ans as brainliest but if you think it has really helped u plz don't forget to thank me....

3 0

3 years ago

One of your delivery trucks traveled 1,200 miles on 55 gallons of gas. How many miles per gallon did the truck get? (Round off y

Aleksandr-060686 [28]

The word "Per" means divide

"miles per gallon" is the same as "miles / gallon"

The truck went 1,200 miles
on 55 gallons

1,200 ÷ 55 = 21.81

7 0

3 years ago

Read 2 more answers

A rocket initially at rest accelerates at a rate of 99. 0 meters/second2. Calculate the distance covered by the rocket if it att

creativ13 [48]

The rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s. Acceleration can be defined as the change in velocity.

<h3>What is acceleration?</h3>

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

$D = v_{t} +\dfrac 12at^2$

Where,

$D$ - final velocity = 445 m/s

$v_0$ - initial valocity = 0 m/s

$a$ - acceleration = 99. 0 m/s²

$t$ - time = 4. 50 s

Put the values in the formula,

$D = 0\times ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times 10^3\rm \ m$

Therefore, the rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s.

Learn more about Acceleration :

brainly.com/question/2697545

7 0

3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

3 years ago