A 1.0 kg ball falls from rest a distance of 19.6 m.
- What was its change in potential energy?
- What was its velocity after falling 19.6 m?

Answer:

the correct  is D

Explanation:

In the calorimetry exercises, if there are no losses, we assume that all the heat lost by one body is absorbed by the other, but if there are losses

         Q_ {yielded} = Q_ {absent} + Q_ {lost}

Since energy cannot be created or destroyed, they only ask to change, therefore the term of the absorbed heat must decrease, consequently the final temperature of the system is lower when there is loss of energy.

When checking the answers, the correct one is D

Answer:

Your ans is B)

Explanation:

Hope its right

if wrong firgive me

Answer:

1)     R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2

Explanation:

1) The resistance of a wire is given by the expression

          R = [tex]\rho \ \frac{l}{A}[/tex]

where ρ is the resistivity of the material, l the length of the wire and A the area of ​​the wire

The area is given by

          A = π r² = π d² / 4

we substitute

         R = (ρ 4 /π)  [tex]\frac{l}{d^2}[/tex]

the amount in parentheses is constant for this case

let's analyze the situation presented, to find the resistance of each wire

* indicate l₁ = l₂ and d₂ = 2 d₁

the resistance of wire 1 is

          R₁ = (ρ 4 /π)  [tex]\frac{l_1}{d_1^2}[/tex]  

the resistance of wire 2 is

          R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}

          R₂ = (ρ 4 /π)   [tex]\frac{l_1}{ (2 d_1)^2}[/tex]

          R₂ = (ρ 4 /π ) [tex]\frac{l_1}{d_1^2}[/tex]   ¼

          R₂ = ¼ R₁

* indicate that d₂ = d₃    and  l₃ = 2 l₂

           R2 = (ρ 4 /π)  [tex]\frac{l_2}{d_2^2}[/tex]

the resistance of wire 3 is substituting the indicated condition

           R3 = (ρ 4 /π 2)  \frac{l_3}{d_3^2}

            R3 = (ρ 4 /π)   [tex]\frac{2 \ l_2}{d_2}[/tex]

            R3 = 2 R₂

let's write the relations obtained

          R₁ = (ρ 4 /π)  [tex]\frac{I_1}{ d_1^2}[/tex]

          R₂ = ¼ R₁

          R₃ = 2 R₂

let's write everything as a function of R1

           R₁ =(ρ 4 /π)   [tex]\frac{l_1}{d_1^2}[/tex]

           R₂ = ¼ R₁

           R₃ = ½ R₁

the resistance of the wire in decreasing order is

           R₁ > R₃ > R₂

2) The power dissipated by a wire is

           P = V I

the voltage is

           V = I R

            I = V / R

substituting

          P = V² / R

therefore the power dissipated by each wire is

wire 1

           P₁ = V² / R₁

wire 2

           P₂ = V² / R₂

           P₂ = [tex]\frac{V^2}{ \frac{1}{4} R_1}[/tex]

           P₂ = 4 P₁

wire 3

           P₃ = V² / R₃

           P₃ = [tex]\frac{V^2}{ \frac{1}{2} R_1}[/tex]

           P₃ = 2 P₁

Therefore, the wire that dissipates the most is wire 2

Answer:

your

answer is

12.004

Answer:

Magnetic domains

Explanation:

Because when the electrons spin around the necleus of an atom, it causes the atom to become a tiny magnet

Answer:

I'm pretty sure its the 2nd one.

Answer:

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.6 x 0   +    0.8 x 0       =   1.6 x v₁     +  0.8 x 3

0 = 1.6v₁  + 2.4

-1.6v₁ = 2.4

v₁  = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

Answer:

7i, 5j, 0k

Explanation:

Add up the two vectors.

Add

i+i       1+6  = 7i

j+j       2+3 = 5j

k+k    -2+2 = 0k

therefore A+B = (7i, 5j, 0k)

10m/s East

Explanation:

p=mv

m1v1 = m2v

p = m1v1

2000kg*(20m/s) = 40000kg*m/s

Add the two masses together for m2

p = m2v2

40000kg*m/s = 4000kg * v2

v2 = 10m/s

Uhm.. based on what? Can you please specify/add an image or some more text and elaborate?

Explanation:

Given that,

We need to find the orbital speed and period of a satellite in orbit [tex]1.44\times 10^2\ m[/tex] above Earth.

The formula for the orbital speed is given by :

[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times5.97\times 10^{24}}{(1.44\times 10^2+6.38\times 10^6)^2}} \\\\v=3.12\ m/s[/tex]

Let T be the time period of the satellite. It can b solved using Kepler's third law i.e.

[tex]T^2=\dfrac{4\pi^2}{GM}r^3\\\\T^2=\dfrac{4\pi^2\times (1.44\times 10^2)^3}{6.67\times 10^{-11}\times 5.97\times 10^{24}}\\\\T=5.44\times 10^{-4}\ s[/tex]

Hence, this is the required solution.

The orbital speed of the satellite is 3.13 m/s and the period of the satellite in orbit is 5.44 × 10⁻⁴ s.

The orbital speed of an astronomical object in a gravitational system is the speed at which it revolves around the central point or the speed required to keep an artificial or natural satellite in orbit.

It can be expressed by using the formula:

[tex]\mathbf{v = \sqrt{\dfrac{GM}{r}}}[/tex]

where;

[tex]\mathbf{v = \sqrt{\dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(1.44 \times 10^2 + 6.38 \times 10^6)^2}}}[/tex]

v = 3.13 m/s

In planetary motion, Kepler's third law states that the square of the period of a planet varies directly proportional to the cubes of the planets' mean distances (d).

The period of the satellite can be estimated by using Kepler's third law. It can be computed by using the formula:

[tex]\mathbf{T^2 =\dfrac{4 \pi^2}{GM}\times r^3}[/tex]

[tex]\mathbf{T^2 =\dfrac{4 \pi^2}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}\times (1.44 \times 10^2)^3}[/tex]

[tex]\mathbf{T^2 =2.96037718\times 10^{-7}}[/tex]

[tex]\mathbf{T =5.44 \times 10^{-4} \ s}[/tex]

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Answer:b yuh

Explanation:

Yuh

Solution :

Given :

Weight = 4 lb

Stretched of a spring in  equilibrium,Δ = 2 ft

a). We know that

F = kΔ

4 = k x 2

k = 2 lb/ft

∴ Stiffness of the spring is, k = 2 lb/ft

Motion is critically damped.

[tex]$\epsilon = 1$[/tex]

[tex]$\epsilon=\frac{C}{C_c} = 1$[/tex]

[tex]$C=C_c[/tex]

[tex]$C=C_c=\sqrt{4k_m}[/tex]

[tex]$C=\sqrt{4 \times 2 \times \left(\frac{4}{32.174}\right)}[/tex]

C = 0.99729 lb.s/ft.

b). Displacement

   [tex]$y(t)=y_0 \sin (\omega t)$[/tex]

[tex]$\dot{y}(t)= y_0 \omega \cos (\omega t)$[/tex]

[tex]$[\dot{y}(t)]_{mon}= y_0 \omega $[/tex]

Initial displacement = 6 inches

    [tex]$y_0 = 0.5 \ ft , \ \ \ \dot y (t) = 3 \ ft/sec$[/tex]

  [tex]$\Rightarrow 3=0.5 \times \omega$[/tex]

 [tex]$\Rightarrow \omega = 6$[/tex]

Therefore, displacement :

    [tex]$y(t)=y_0 \sin (\omega t)$[/tex]

   [tex]$y(t)=0.5 \sin (6 t)$[/tex]

Answer:

Electricity ⚡️

Explanation:

molecular weights are written in the picture.

CH4<NH3<H2O<Cl2

Answer:

electric charge

Answer:

Electric charge.

Explanation:

Coulomb, unit of electric charge in the metre - kilogram - second - ampere system, the basis of the SI system of physical units. It is abbreviated as C. The coulomb is defined as the quantity of electricity transported in one second by a current of one ampere.

Answer:

40m/s

Explanation:

Answer:

8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2

Explanation:

Exact number is 8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2

Answer:

B radiation

Explanation:

every type of transmitted waves are radioactive in a way.

Answer: D. When both objects reach the same temperature.

Explanation: When will heat flow between the objects stop? Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.

Answer: Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.

so the answer to your question is c

Explanation:

Answer:

a. increases

Explanation:

the law says

R=ρ[tex]\frac{L}{A}[/tex]

R= resistance

ρ= permiability of wire

L= length of the wire

A= area of the wire

Answer:

What is the first velocity of the car with three washers

at the 0.25 meter mark? 0.19

What is the second velocity of the car with three

washers at the 0.50 meter mark? 0.45

Explanation:

Look at the question carefully.

The person above me got the answer wrong.

That answer is for a very similar question, but not this one.

(1) The first velocity of the car with two washers at the 0.25 meter mark is 0.125 m/s.

(2) The second velocity of the car with two washers at the 0.5 meter mark is 0.25 m/s.

Velocity is the rate of change of displacement with time. The velocity of the  cars depends on displacement and time of motion.

First velocity of the car at the 0.25 meter mark

The first velocity is calculated as follows;

v1 = 0.25 m / t1

let t₁ = 2 s

v₁ = 0.25/2

v₁ = 0.125 m/s

Second velocity of the car at the 0.5 meter mark

v2 = 0.25 m / (t2 – t1)

let t₂ = 3 s

v₂ = 0.25(3 - 2)

v₂ = 0.25 m/s

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here, d is the distance covered.

And t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

Here, t1 is the average time for the first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

Therefore, here, t2 is the average time for the second distance.

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Answer:

The Cell Membrane is found in both plant and animal cells.

Explanation:

Chloroplast, Large Vacuole and Cell Wall are all found in plant cells.

Answer:

The water and peanut butter have different chemical properties.

(Lipids are not soluble in water)

Answer:

a. The water and peanut butter have different chemical properties.

Explanation:

Answer:

Explanation:

B. Central Nervous System

Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.

When a tennis racket hits a ball, the ball exerts a force on the racket, and according to Newton's third law of motion, the racket exerts an equal and opposite force on the ball. This means that the ball pushes back on the racket in the opposite direction to which the racket struck the ball.

This principle is often referred to as "action-reaction" or "equal and opposite forces." When the racket collides with the ball, the force applied by the racket causes the ball to accelerate in the opposite direction, leading to its movement away from the racket.

Therefore, when a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.

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