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2 years ago

11

Martha is cycling at a speed of 20 kilometers per hour. How long will it take her to cover a distance of 60 kilometers? speed=di

stancetime

2 answers:

ser-zykov [4K]2 years ago

7 0

It will take 3 hours hope it helped :D

aivan3 [116]2 years ago

6 0

It will take 3 hours

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In a related practical activity, 2 litres of a gas were heated from 35 °C to 75 °C. If the pressure was kept constant, calculate

Pachacha [2.7K]

Answer:

2.26l

Explanation:

From the general gas equation:

P1V1/T1 = P2V2/T2

Since pressure remained constant we can say:

V1/T1 = V2/T2

so to convert to kelvin add 273 to both temperature values then we can say:

1 m^3= 1000 L

2l=0.002m^3

Then;

0.002/308=V/348

V=(0.002/308)348

Final volume=0.002259m^3

=2.6l(1 decimal place)

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1 year ago

A Martian rover is descending a hill sloped at 30° with the horizontal. It travels with a constant velocity of 2 meters/second.

kodGreya [7K]

Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.

8 0

3 years ago

Read 2 more answers

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

2 years ago

The graph below show how natural processes and human activities affect climate

Kaylis [27]

Is it just me or can anyone else not see a graph? (Because I can’t)

8 0

3 years ago

Read 2 more answers

A marble resting near the edge of .90 m high table is given an initial horizontal speed of 1.24 m/s. What will be it’s horizonta

Maru [420]

Answer:

0.53 m

Explanation:

First of all, we have to consider the vertical motion of the ball, in order to find the time it takes for the marble to reach the ground. The initial height is $h=0.90 m$ , the initial vertical velocity is zero, while the acceleration is $g=-9.81 m/s^2$ , so the vertical position at time t is given by

$y(t)=h-\frac{1}{2}gt^2$

By demanding y(t)=0, we find the time t at which the ball reaches the ground:

$0=h-\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(0.9 m)}{9.81 m/s^2}}=0.43 s$

Now we can find the horizontal range of the marble: we know the initial horizontal speed (v=1.24 m/s), we know the total time of the motion (t=0.43 s), and since the horizontal speed is constant, the total distance traveled on the horizontal direction is

$x=vt=(1.24 m/s)(0.43 s)=0.53 m$

8 0

3 years ago