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olya-2409 [2.1K]
3 years ago
6

A short, thick wire is the best conductor. True or false?

Physics
2 answers:
Firdavs [7]3 years ago
7 0
I believe that the answer is true!
svlad2 [7]3 years ago
3 0
True












BcFbBcFbHFfnFNjFhrRjThyjtzfbFBfbvhhhfFbBfFFBbffb
I did that BC I got frustrated sorry
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a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
GalinKa [24]

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

∴acceleration, a= 14/70 = 0.2 m/s²

now on applying conservation of linear momentum

pi=pf            pi=0 (initially at rest)

0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s

v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

0- (0.3) ²= -2 x 0.2 x 5

s= 0.09/0.4 ≈ 0.3 m

Learn more about distance here-

brainly.com/question/15172156

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6 0
2 years ago
A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o
Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

KE_i=\dfrac{1}{2}mu^2

KE_i=\dfrac{1}{2}0.3\times 10^2

KEi= 15 J

Final kinetic energy

KE_f=\dfrac{1}{2}mv^2

KE_f=\dfrac{1}{2}0.3\times 5^2

KEf=3.75 J

The  energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

 V=\sqrt{r.g}

Now by putting the values

V=\sqrt{0.8\times 10}

V= 2.82 m/s

So kinetic energy KE

KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

                                                                                                    = 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0
3 years ago
What does it mean that are bodies are in osmotic equilibrium but chemical and electrical disequilibrium? Why? What molecules are
Whitepunk [10]

Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

8 0
3 years ago
Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t
Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0
3 years ago
The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em
Vilka [71]

Answer:

\rho=995.50\ kg.m^{-3}

\bar w=9765.887\ N.m^{-3}

s=0.9955

Explanation:

Given:

  • volume of liquid content in the can, v_l=0.355\ L=3.55\times 10^{-4}\ L
  • mass of filled can, m_f=0.369\ kg
  • weight of empty can, w_c=0.153\ N

<u>So, mass of the empty can:</u>

m_c=\frac{w_c}{g}

m_c=\frac{0.153}{9.81}

m_c=0.015596\ kg

<u>Hence the mass of liquid(soda):</u>

m_l=m_f-m_c

m_l=0.369-0.015596

m_l=0.3534\ kg

<u>Therefore the density of liquid soda:</u>

\rho=\frac{m_l}{v_l} (as density is given as mass per unit volume of the substance)

\rho=\frac{0.3534}{3.55\times 10^{-4}}

\rho=995.50\ kg.m^{-3}

<u>Specific weight of the liquid soda:</u>

\bar w=\frac{m_l.g}{v_l}=\rho.g

\bar w=995.5\times 9.81

\bar w=9765.887\ N.m^{-3}

Specific gravity is the density of the substance to the density of water:

s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

3 0
3 years ago
Read 2 more answers
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