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larisa [96]
2 years ago
5

A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma

gnitude of the vertical component of its initial velocity?
Physics
2 answers:
tangare [24]2 years ago
5 0

Answer:

8.7

Explanation:

(10)sin60

Flura [38]2 years ago
3 0

Answer:

5 sq. root 3

Explanation:

theta= 60°

=> u sin theta = 10 × sin 60

= 10× sq. root 3/2

= 5 sq. root 3

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An occluded front forms when a cold front catches up with a warm front. So it would bring Sun and warmth.
 
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3 years ago
two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot
Iteru [2.4K]

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

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3 years ago
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7 0
3 years ago
If two 100 ohms resistors are placed in series, their total resistance is what?
Sindrei [870]

Answer:

add the Resistance

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3 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

So due to wind time will be more

4 0
3 years ago
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