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Elodia [21]
3 years ago
14

The ability to speak does not say you intelligent essay​

Physics
1 answer:
Bad White [126]3 years ago
6 0

Answer:

Change is for better and is necessary too. Everyone knows it but only the ones who apply it are the ones whom we call as intelligent. When a person changes in an order to make his life better, he chooses the right thing because staying put at a place in life doesn't take anyone anywhere.

Taking the example of a few infamous personalities, whom we consider as intelligent, we see a common thing that each one of them had a certain vision in life and in order to achieve it, they were ready to accept any kind of change and still are. So, this makes it clear that change is what leads us to our goal and in order to achieve it, we need to change with time.

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A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring
borishaifa [10]

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2  = 0 m/s

putting all value in the above equation

1.14 *2.0+ 0 =(1.14+1.14)v^{2}

v =\frac{1.14*2.0}{1.14+1.14}

v = 1 m/s

4 0
3 years ago
A car's gas tank contains 58.7 kg
Bogdan [553]

Answer:

721 kg/m^3

Explanation:

Trust me bro

8 0
3 years ago
A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0
3 years ago
Read 2 more answers
What is Nuclear energy
Yuliya22 [10]

Answer:

Nuclear energy is that released by dividing the nuclei of heavy atoms (also called under the name Nuclear Fission)

Explanation:

In this process, a large amount of heat is generated that can be used to obtain mechanical energy, which is used to generate electrical energy.

Nuclear energy is a sustainable source of energy having a low impact on the environment.

3 0
3 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
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