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Bond [772]
3 years ago
10

A cheetah can run at 30 m/s but only for about 12s. how far will it run in that time

Physics
2 answers:
Mashutka [201]3 years ago
6 0
About 360 meters I hope this helps 
Anika [276]3 years ago
3 0

Answer: 360 meters

Explanation: A cheetah can run at 30m/s, this means that in each second the animal can travel a distance of 30m.

If the cheeta runs for 12 seconds at that speed, then it travels 12 times that distance.

The distance that a cheetah that runs at 30m/s by 12 seconds is:

D = 30m/s*12 = 360 m

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Which two of the following reduce the background noise in magneto- medicine?
horrorfan [7]

Answer:

The answer is b.) and d.)

Explanation:

The options to reduce the background noise in magneto-medicine are given as follows:

a.) Orienting the heart parallel to the Earth's field

   - This will have no significant effect on the measurements.

b.) Taking the difference of two nearby sensor measurements (gradiometer).

    -This answer is TRUE.

c.) Placing the heart in a perpendicular fashion to the Earth's magnetic field.

   _ This answer also will not have any significant effect on measurements.

d.) Using physical means to shield environmental fields.

   - This answer is TRUE.

4 0
3 years ago
An object accelerates 12.0 m/s² when a force of 6.0 newtons is applied to it.
Nimfa-mama [501]

Answer:

<h2>0.5 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{6}{12}  =  \frac{1}{2}  \\

We have the final answer as

<h3>0.5 kg</h3>

Hope this helps you

7 0
3 years ago
Read 2 more answers
What term is used for a weather condition characterized by lack of precipitation over a long period of time?
Marrrta [24]

Answer:

Explanation:

<h2><u><em>Drought</em></u></h2>
4 0
3 years ago
Read 2 more answers
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0
3 years ago
-------------<br>--------------<br>--------------
dalvyx [7]

Answer:

-------------

--------------

--------------

Explanation:

5 0
4 years ago
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