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Tresset [83]
2 years ago
8

for an object that is speeding up in the positive direction what does the displacement vs. time graph look like

Physics
1 answer:
avanturin [10]2 years ago
3 0

For an object that is speeding up in the positive direction, its positive displacement is greater every second, and the AMOUNT greater every second is greater than the AMOUNT greater was in the previous second.

So the graph of displacement vs. time is rising as time goes on, and the rise is becoming steeper as time goes on.

The graph is <em>curving upward</em> as time goes on.

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90 kilometers an hour

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3 0
2 years ago
PLEASE HELP!
Harman [31]
Hey user!

your answer is here..

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8 0
3 years ago
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.
rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = \frac{1}{2}m_{p}v²

v = √( 2K / m_{p} )

lets relate the cross-sectional area A of the beam to its diameter D;

A = \frac{1}{4}πD²

now, we substitute for v and A

n = I / \frac{1}{4}πeD² ×√( 2K / m_{p} )

n = 4I/π eD² × √(m_{p} / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n =  1.98695 × 10¹⁸ × 1.6157967  × 10⁻⁵

n = 3.2 × 10¹³ m⁻³  

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0
2 years ago
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