distace x time graph will be ................ if the body is in ununiform motion​

Answer:

Explanation:

Ek=1/2mv^2

m=50+10=60,v=10

Ek=1/2*60*100=3000J

Answer:

       k = 1.30

Explanation:

For this exercise let's write the capacitance in air and with dielectric

air             C₀ = Q / DV

dielectric  C = k Q / DV

They tell us that the capacitor is charged and then the battery is disconnected, therefore the charge stored on the plate remains constant.

therefore the capacitance a changes to the value

           C = k C₀

The voltage in the presence of dielectric must meet the relationship

           ΔV = ΔV₀ / k

           k = ΔV₀ /ΔV

let's calculate

           k = 60/46

           k = 1.30

Answer:

B radiation

Explanation:

every type of transmitted waves are radioactive in a way.

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:

Explanation:

B. Central Nervous System

Answer:

Your ans is B)

Explanation:

Hope its right

if wrong firgive me

Answer:

(a) The reduced pressure is 0.2711 MPa

The reduced temperature is 2.54 K

(b) The initial volume of the piston is approximately 0.806 m³

(c) The mass of CO₂ is approximately 16.9884 kg

(d) The work done, W is approximately 388.023 kJ

(e) The heat transfer is approximately -2,650.1904 kJ

Explanation:

The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K

The initial pressure of the gas, P₁ = 2 MPa

The final temperature of the gas, T₂ = 350°C

The final volume of the gas = 1 m³

(a) For an isobaric process, we have;

The reduced pressure,

[tex]P_r = \dfrac{P}{P_c}[/tex]

The critical pressure of carbon dioxide, [tex]P_c[/tex] = 7.3773 MPa

[tex]P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa[/tex]

The reduced pressure, [tex]P_r[/tex] = 0.2711 MPa

The critical temperature, [tex]T_c[/tex] = 304.13 K

The reduced temperature, [tex]T_r[/tex], is given by the following formula;

[tex]T_r = \dfrac{T}{T_c}[/tex]

Therefore, [tex]T_r[/tex] = (773.15 K)/(304.13 K) = 2.54216947 K

The reduced temperature, [tex]T_r[/tex] ≈ 2.54 K

(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁

∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³  ≈ 0.806 m³

The initial volume of the piston, V₁ ≈ 0.806 m³

(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;

n = P·V/(T·R)

The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles

The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg

(d) The work done, W = P·([tex]V_f - V_i[/tex])

W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227

The work done, W ≈ 388.023 kJ

(e) The heat transfer dQ = m·[tex]c_p[/tex] ×(T₂ - T₁)

[tex]c_p[/tex] for CO₂ ≈ 1.04 kJ/(kg·K)

∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ

Therefore, the heat transfer = dQ = -2,650.1904 kJ

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.

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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Answer:

40m/s

Explanation:

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:

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:

I think it's because the light waves travel faster than the sound waves.

The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.

Light occurs in the electromagnetic spectrum. Recall that light can be transmitted through vaccuum unlike sound.

The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.

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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:

s=1/2at^2

s=1/2 x 4 x(10)^2=200m

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

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:

Electricity ⚡️

Explanation:

Answer:b yuh

Explanation:

Yuh

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

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:

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:

Explanation:

mass (m) = 65 kg

velocity (v) = 4.6 m/s

Kinetic energy (KE)

= 1/2 * m * v²

= 1/2 * 65 * 4.6²

=  687.7 J

hope it helps :)

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:

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:

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: