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saul85 [17]
3 years ago
5

Help awnser need experts​

Physics
1 answer:
pishuonlain [190]3 years ago
5 0

12. The answer would be C. 1.50 s. This is because if you divide 60 by 40, you will get 1.5.

13. For this one I'm not sure, but what I can tell you is that the heavier something is the faster it will sink, the lighter it is, it will float.

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Help me! Btw he’s playing golf.
fiasKO [112]

Answer:

Muscular energy

Explanation:

Hope it helps!!!!

5 0
3 years ago
A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what a
Aneli [31]

Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

θ = (225 - 180) ± arcsin(13/30)

θ = 45 ± 25.679...

70.679 ≤ θ ≤ 19.321

components of vector B would be

Bx = |B|cosθ

By = |B|sinθ

3 0
2 years ago
A spindle connects two wheels with fixed axles by a fan belt. If the effort wheel is larger than the resistance wheel, which axl
MariettaO [177]
Power = Iω (constant) as they are connected together, since effort axle has large radius than resistance axle, so moment of inertia of effort axle is also more as compared to resistance axle, so angular speed of effort axle is less than the resistance axle. So answer is B. resistance axle  will have more angular speed as its moment of inertia is less for the same power.
8 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
Electrically charged particles are found primarily in
lesantik [10]

Answer:

the ionosphere

.......................................................

8 0
2 years ago
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