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notsponge [240]
3 years ago
15

A body is accelerated continuously. What is the form of the graph?

Physics
1 answer:
Inga [223]3 years ago
6 0
That depends on what quantity is graphed.

It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.

-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces. 
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.

-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.

-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.

-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line. 
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks. 
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.

-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves.  Each piece curves up if speed is increasing, or down if
speed is decreasing.

-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.

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If planet A is three times as far from planet C, then the period of its orbit will be __ times as long
liubo4ka [24]
I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em>  5.2</em>  times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.


6 0
3 years ago
8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow
Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

8 0
3 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
A baseball thrown from the outfield to home plate does not have which of the following types of energy while it's in the air? A.
posledela
The ball does not have D. Radiant energy.
7 0
3 years ago
A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T
mart [117]
<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

                          = 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

  = √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E =  4.32 × 10^-13 Joules

  r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

                            = 2.775 × 10^7 rad /sec

3 0
3 years ago
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