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s344n2d4d5 [400]
3 years ago
5

A 57 kg go-kart is pushed with a force of 15 N. What will its acceleration be?

Physics
1 answer:
Zanzabum3 years ago
6 0
F=ma is newton's second law. a=F/m. F= 15, m = 57 .... a = 15/57 about 1/4 m/s per second. 0.25 ... 
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A pelican flying along a horizontal path drops
Ludmilka [50]

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s

The initial horizontal speed of the pelican is given by;

X = vₓt

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vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

8 0
3 years ago
When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored
adoni [48]

The energy stored in a capacitor is

E = (1/2) · (capacitance) · (voltage)²

E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

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(That's 0.000432 of a Joule)

4 0
3 years ago
The modern periodic table has______ groups. (Write your answer as a number.)
kobusy [5.1K]
It contains 18 groups
4 0
3 years ago
Read 2 more answers
the acceleration due to gravity on earth is 9.80 m/s2. if the mass of a giraffe is 1,470 kg, what is the weight of the giraffe?
Brrunno [24]
Weight  = Mass * gravity

             =  1470* 9.8 = 14406 N  ≈ 14,400 N
5 0
3 years ago
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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
3 years ago
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