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3 years ago

How could you figure out how fast you can run

Physics

2 answers:

Arlecino [84]3 years ago

7 0

Well you first need an equation, then put it on a graph

WARRIOR [948]3 years ago

6 0

Well have someone time you on a sprint for as many minutes as they want and when they stop the stopwatch u can see how fast you run for example they can see how fast u run in 5 minutes which is 2 minutes just and example Hope this helps u:)

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Tick (3) the correct statement about electrostatic charges.

Radda [10]

Answer:

similar type of electric charges attract one another

I think this is a coorect staement

6 0

3 years ago

Read 2 more answers

The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle

Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0

3 years ago

Can someone help me?

Leviafan [203]

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0

3 years ago

20.0 Ω resistor and a 2.50 μF capacitor are connected in parallel with signal generator. The signal generator produces a sinusoi

Free_Kalibri [48]

Answer:

0.15A

Explanation:

The parameters given are;

R=20.0 Ω

C= 2.50 μF

V= 3.00 V

f= 2.48×10^-3 Hz

Xc= 1/2πFc

Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6

Xc= 25666824.1

Z= 1/√(1/R)^2 +(1/Xc)^2

Z= 1/√[(1/20)^2 +(1/25666824.1)^2]

Z= 1/√(2.5×10^-3) + (1.5×10^-15)

Z= 20 Ω

But

V=IZ

Where;

V= voltage

I= current

Z= impedance

I= V/Z

I= 3.00/20

I= 0.15A

6 0

3 years ago

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

Elanso [62]

On Earth, the period of a pendulum is given by:
$T_{earth}=2\pi \sqrt{ \frac{L}{g_{earth} }$
where L is the length of the pendulum and $g_{earth}=9.81~m/s^2$ is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
$T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }$
where $g_{mars}=3.71~m/s^2$ is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
$\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} } = \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63$
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.

6 0

4 years ago