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larisa86 [58]
3 years ago
14

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

for 1.70 seconds, at which point its fuel abruptly runs out. air resistance has no effect on its flight. what maximum altitude (above the ground) will the rocket reach?
Physics
1 answer:
Harman [31]3 years ago
5 0
<span>When the fuel  of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction. 

Given:
a = 86 m/s^2 
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2) 
d = (0) (1.7) + 0.5 (86) (1.7)^2 
d = 124.27 m 

vf = vi + at 
vf = 0 + 86 (1.7) 
vf = 146.2 m/s (velocity when the fuel is consumed completely) 

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at 
0 = 146.2 + (-9.8) (t) 
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2) 
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2) 
d = 1090.53  m

Then, we determine the maximum altitude:
 d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
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A: the intensity

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Does passing a magnet through a coil of wire break off it’s electric current
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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

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3 years ago
A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca
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Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

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a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

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