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rosijanka [135]

3 years ago

7

Which three terms are needed to describe the energy a BASE Jumper has as

1 answer:

STatiana [176]3 years ago

4 0

A Potential (as she is off the ground she has gravitational potential energy)
B kinetic (because she’s moving down)
and
C gravitational (as gravity is pulling her back down)

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If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w

DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

$d = 2\pi r$

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

$d = 2\pi r = 2\pi*6371 km = 40030.2 km$

Now, if we divide the above distance by the speed of the car we can find the time:

$t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d$

Therefore, Scobie will take 10 days to drive around Earth's equator.

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0

3 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

c=1

So

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

3 years ago

Read 2 more answers

Which part of the battery is electron enriched?

dexar [7]

The answer is b. Negative terminal.

5 0

3 years ago

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Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter

Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

$\Delta U= Q-W$

where

$\Delta U$ is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

$\Delta U=U_f-U_i=134 J-87 J=47 J$

While the heat added to the system is

$Q=500 J$

therefore, the work done by the system is

$W=Q-\Delta U=500 J-47 J=453 J$

5 0

3 years ago

Read 2 more answers

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Yuri [45]

P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]

3 0

3 years ago