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Inessa05 [86]
3 years ago
13

How much meters is a mile

Physics
1 answer:
elena-s [515]3 years ago
6 0

Roughly 1609 meters in one mile

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A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?
Kobotan [32]

In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s<span>^2</span>

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3 years ago
What would be the effect on the annual march of the seasons if earth's axis were not inclined relative to the plane of the eclip
victus00 [196]
The poles wouldn't get light all year round.
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3 years ago
A parallel circuit contains ___________.
kotykmax [81]
The answer to your question is D
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3 years ago
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Which of the following is the smallest in mass?
vodka [1.7K]

The correct choice is

D. electron

mass of proton is 1.67 x 10⁻²⁷ kg and has positive charge on it.

mass of neutron is slightly greater than the mass of proton. the mass of neutron is 1.675 x 10⁻²⁷ kg and has no charge on it.

mass of electron is 9.31 x 10⁻³¹ kg

comparing the masses of electron, proton, neutron and the nucleus, we see that the mass of electron is smallest.

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A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass
Talja [164]

Answer:

a)1.93 kg-m^2

b) 1.45  kg-m^2

c) = 0

d) 1.15 kg-m^2

Explanation:

mass of the bar M = 4 kg

length of the bar = 2 m

mass of balls m1= m2= 0.3 kg

moment of inertia of bar I= \frac{ML^2}{12}

about an axis perpendicular to the bar through its center.

a) MOI of bar + 2×m×(L/2)^2

I= \frac{ML^2}{12}+ 2m\frac{L}{2}^2

now putting the values of m, M and L as above and solving we get

I= 1.93 kg-m^2

b) perpendicular to the bar through one of the balls

I=M\frac{L^2}{3} +mL^2

I=4\frac{2^2}{3} +0.3\times4^2= 1.45  kg-m^2

c) parallel to the bar through both balls

zero as the no mass distribution along the parallel to the bar through both balls.

d) parallel to the bar and 0.500 m from it.

I=(M+2m_1)\frac{1}{2}^2

putting values and solving we get

1.15 kg-m^2

4 0
4 years ago
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