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sergey [27]
2 years ago
7

A car is traveling 50 mph. It leaves Boston, Massachusetts at 5 a.m. and is

Physics
1 answer:
aleksandrvk [35]2 years ago
6 0
7 A.M
I’m pretty sure that’s it
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the coefficient of static friction between a 40 kg picnic table and the ground below is .43. what is the greatest horizontal for
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The force equals the coefficient of static friction times the weight. Use gravity g=9.8 m/s^2
0.43*40*9.8=16.856 N
4 0
3 years ago
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S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to
denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0
1 year ago
Leo la conclusión e identificó dos situaciones que causan que una lengua desaparezca buscó otra situación que también incide al
Novosadov [1.4K]

Answer:

yes smh

Explanation:

4 0
3 years ago
Control rods in a nuclear reactor are used to:
kiruha [24]
B.absorb neutrons to prevent chain reactions which become uncontrollable
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3 years ago
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Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done
blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=\frac{F}{m}

Hence acceleration of the block is given by:

\Rightarrow a=\frac{25}{2}=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+\frac{1}{2}at^2

Substitute the values

\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2

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Now work done is given by:

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W =1406.25 J

3 0
3 years ago
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