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IrinaK [193]
3 years ago
10

You run from your friends house that is 1k away. You then walk home. What distance did you travel?

Physics
1 answer:
kolezko [41]3 years ago
6 0

Would would have moved a total distance of 2k, because walking is a slower speed does not change distance

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If one 9V battery is used in a circuit with a total resistance of 39Ω, what is the current in the circuit?
Ket [755]

Answer:

Using V= IR

I= 0.2307 Ampere

6 0
2 years ago
A falcon can descend with a speed 250 km/h. If a falcon flies at this speed for 2.0 s and then flies a 100 m in 2.5 s, what is t
Vadim26 [7]

Answer:

v= s/t

Explanation:

250 km/ h =69.44m/s

S1=2 times 69.44 ≈ 139m

Next 2.5 seconds:

S2 = 100m

Average speed:

v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h

3 0
2 years ago
The tension in the horizontal towrope pulling a water-skier is 250 N while the skier moves due west a distance of 50 m. How much
icang [17]

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3 years ago
Antonina throws a coin straight up from a height of
vichka [17]

Answer:

s=vt-\frac{1}{2}gt^2

Explanation:

We could use the following suvat equation:

s=vt-\frac{1}{2}gt^2

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

g=-9.8 m/s^2

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s

where the negative sign means the direction is downward.

5 0
3 years ago
A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of r
Sedbober [7]

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

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Given the data in the question;

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Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

t_{min = ( λ / 4n )

so we simply substitute in our given values;

t_{min = ( 463 × 10⁻⁹ m ) / 4(1.35)

t_{min = ( 463 × 10⁻⁹ m ) / 5.4

t_{min = ( 463 × 10⁻⁹ m ) / 4(1.35)

t_{min = 8.574 × 10⁻⁸ m

t_{min = 85.74 × 10⁻⁹ m

t_{min = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

5 0
3 years ago
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