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3 years ago

What is the magnitude of the orbital velocity of the earth in m/s?

1 answer:

8 0

Circumference C=2πr
C=2π(1.5x10^8)=9.42x10^8 

In 365 Days there are 8760hr 

V=distance/time 

V=(9.42x10^8)/8760=107534.2km/hr

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An object is moving initially with a velocity of 4.7 m/s . After 3.9 s the object's velocity is -2.1 m/s . What is the object's

IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

8 0

3 years ago

How to make a paper airplane curve down and fly upside down?

Ainat [17]

Throw it sideways and try to make it spin around but it needs to be thrown high up then it should kinda glide down

8 0

2 years ago

Read 2 more answers

What’s the unit for velocity?

torisob [31]

Answer:

meter per second

Explanation:

It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.

Examples

feet per second

miles per hour

4 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

2 years ago

If I accidentally touch a stove and burn myself, which type of heat transfer is occurring?

allochka39001 [22]

Answer:

radiation is the correct answer

5 0

3 years ago