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3 years ago

PLEASE HELP

2 answers:

4 0

Don't over think it!

For example, if you do 5 laps in 5 minutes then you do one lap per minute. If you use that example and you apply it to your problem you'll notice that the answer is simply C. 1km/min.

Best wishes!

Grace [21]3 years ago

4 0

C. 1km/min because the car covers a distance of 5km over a period of 5 minutes, so the car travels 1 kilometer each minute.

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The unit of electric potential or electromotive force is the

Natalka [10]

That would be called VOLT

4 0

3 years ago

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle $F_{net}$ is given by $F_{net}=F-f$

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for $F_{net}$

$F_{net}=ma$ hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

6 0

3 years ago

What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.

7 0

3 years ago

For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas

nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

$F=\dfrac{mv^2}{R}$ , m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

$F=\dfrac{(16)^2m}{60}=4.26m\ N$

Case 2.

R = 15 m v = 8 m/s

$F=\dfrac{(8)^2m}{15}=4.26m\ N$

Case 3.

R = 30 m v = 4 m/s

$F=\dfrac{(4)^2m}{30}=0.54m\ N$

Case 4.

R = 45 m v = 4 m/s

$F=\dfrac{(4)^2m}{45}=0.36m\ N$

Case 5.

R = 30 m v = 16 m/s

$F=\dfrac{(16)^2m}{30}=8.54m\ N$

Case 6.

R = 15 m v =12 m/s

$F=\dfrac{(12)^2m}{15}=9.6m\ N$

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0

2 years ago

A proton has been accelerated from rest through a potential difference of -1000 v . part a what is the proton's kinetic energy,

wlad13 [49]

We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
$\Delta U + \Delta K=0$
which means
$\Delta K = - \Delta U$
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference ( $\Delta V=-1000 V$ ):
$\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV$
Therefore, the kinetic energy gained by the proton is
$\Delta K = -(-1000 eV)=1000 eV$
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>

4 0

3 years ago

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