An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is

Answer:

option (C) is right answer

Explanation:

I think it's helps you

Answer:

Option C is correct

Explanation:

hope it helps you

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

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Answer:

yes

it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

Answer:

its a. and jusing that youl

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: [tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex], where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

[tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}[/tex] = [tex]\frac{81}{41}[/tex] = 1.9756

y-values:

[tex]y_{cm}[/tex] = [tex]\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}[/tex] = [tex]\frac{-90}{41}[/tex] = -2.1951

center of mass:

(1.9756, -2.1951)

Answer:

v = 6.195 m / s

Explanation:

For this exercise we can use the conservation of energy, for the system formed by the ball and the pulley

starting point. Higher before releasing the system

        Em₀ = U = M g h

final point. When the ball has lowered h = 2

        Em_f = K = ½ M v² + ½ I w²

the energy is preserved

        Em₀ = Em_f

        M g h = ½ M v² + ½ I w²

angular and linear velocity are related

         v = w r

        w = v / r

indicate that the moment of inertia is

         I = ½ m r²

we substitute

          M g h = ½ M v² + ½ (½ m r²)  (v/r) ²

          ½ v² (M + [tex]\frac{1}{2}[/tex] m) = M g h

          v² = [tex]2gh \ \frac{M}{M + \frac{m}{2} }[/tex]

let's calculate

          v = [tex]\sqrt{ 2 \ 9.8 \ 2 \ \frac{70}{70 + 1.5} }[/tex]

          v = 6.195 m / s

Answer:

1/Re= 1/R1 + 1/R2

Explanation:

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by  [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and

R₂ = 112 Ω is 0.25 Ω/min

For a given resistor connected in parallel;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making R from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.

So, by differentiating the given equation of the resistor in parallel with respect to time t;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex],   we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min

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Answer:

All of the above.

Explanation:

A person is said to be at rest when he or she is not performing any action. The heartbeat stays normal in this condition.

A person is said to perform a moderate activity when his heart beats faster than the normal rest condition. It requires extra effort within the limit. For example, walking and dancing are examples of moderate activity.

A person is said to perform a vigorous activity when his heart beats much than in moderate activity. It requires a lot of effort over the limit. Running is an example of vigorous activity.

In view of the above information, all the options given in the question are true. Hence, the correct answer will be:

All of the above.

Answer:

The atomic mass of an element is affected by the distribution of its isotopes because each isotope of an element has a different number of neutrons in the nuclei of its atoms.

Explanation:

please give me brainlyiest if its right

Answer:  lose weight at high altitudes.

Explanation:

Answer:

Just a week at high altitudes can cause sustained weight loss, suggesting that a mountain retreat could be a viable strategy for slimming down. Overweight, sedentary people who spent a week at an elevation of 8,700 feet lost weight while eating as much as they wanted and doing no exercise

Answer:

i am guessing for reflection but not so sure

Answer:

okay

Explanation:

please I don't know

Answer: yes

Explanation:

Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.

Answer:

Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Explanation:

P.S - The exact question is -

Given - A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to  [tex]f(t) = 8 cos(4t)[/tex] is applied to the system.

To find - Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.

Proof -

Given that,

Mass = 1 slug

We know that, 1 slug = 32 lb

Now,

Force, f = kx

⇒32 = k(2)

⇒k = 16

Now,

Given that, C = 8 ( 8 times the instantaneous velocity)

Now,

The differential equation of motion is equals to

mx'' + Cx' + kx = 8 cos(4t)

⇒x'' + 8x' + 16x = 8 cos(4t)               ...........(1)

Let the General solution of equation (1) be

x(t) = x(c) + x(p)

Now,

The auxiliary equation is

m² + 8m + 16 = 0

m² + 4m + 4m + 16 = 0

m (m+4) + 4 (m+4) = 0

⇒(m+4)(m+4) = 0

⇒m = -4, -4

So,

The Complimentary equation becomes

x(c) = [tex]Ae^{-4t} + Bte^{-4t}[/tex]                  ...........(2)

Now,

Let the particular solution be

x(p) = C cos(4t) + D sin(4t)

x'(p) = -4C sin(4t) + 4D cos(4t)

x''(p) = -16C cos(4t) - 16D sin(4t)

It also satisfy equation (1)

Equation (1) becomes

-16C cos(4t) - 16D sin(4t) + 8 [ -4C sin(4t) + 4D cos(4t) ] + 16 [ C cos(4t) + D sin(4t) ] = 8 cos(4t)

⇒-16C cos(4t) - 16D sin(4t) - 32C sin(4t) + 32D cos(4t) ] + 16C cos(4t) + 16D sin(4t) ] = 8 cos(4t)

⇒-4C sin(4t) + 4B cos(4t) = cos(4t)

By comparing, we get

4B = 1 , A = 0

⇒ B = [tex]\frac{1}{4}[/tex] , A = 0

So, The particular solution becomes

x(p) =  [tex]\frac{1}{4}[/tex] sin(4t)

Now,

The General solution becomes

x(t) = [tex]Ae^{-4t} + Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)       .......(3)

Now,

Given that, At t = 0, initial velocity is zero and the system starts equilibrium

⇒x(0) = 0, x'(0) = 0

By putting t = 0 in equation (3) , we get

A = 0

Now,

Differentiate equation (3), we get

x'(t) = [tex]-4Ae^{-4t} + Be^{-4t} - 4Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] *4 cos(4t)

Put t = 0, we get

0 = -4A + B + 1

⇒B = -1

∴ we get

The general solution becomes

x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Answer:

b) 1.63x10-9N

Explanation:

i dont think this is right answer

Answer:

C or 8.17x10^-10 N

Explanation:

I was looking for the explanation but the answer is correct.

Answer:

  a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]  ,       t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Explanation:

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

         Wₓ - T = m a

Y axis  

         N- W_y = 0

         N = W_y

for pulley

          ∑τ = I α

           T R = (½ m_p R²) α

let's use trigonometry for the weight components

         sin β = Wₓ / W

         cos β = W_y / W

         Wx = W sin β

angular and linear variables are related

          a = α R

          α = a / R

we substitute and group our equations

         W sin β - T = m a

         T R = ½ m_p R² (a / R)

         W sin β - T = m a

                        T = ½ m_p a

we solve the system of equations

         W sin β = (m + ½ m_p) a

          a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]

let's find the time to travel the distance (d) through the block

          x = v₀ t + ½ a t²

          d = 0 + ½ a t²

          t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Answer:

A

Explanation:

Answer:

can you explain in Hindi language

because i learn hindi

Answer74.3

:

Explanation:

Answer:

Right

Explanation:

Coil move right yes

Answer:We have , a relation in frequency f and wavelength  λ of a wave having the velocity v as ,

           v=fλ ,

given f=60Hz , λ=20m ,

therefore velocity of wave , v=60×20=1200m/s

Answer:

m = 1.31 kg

Explanation:

Given that,

The speed of duck, v = 9.1 m/s

The magnitude of momentum, p = 12 kg-m/s

We need to find the mass of the duck. We know that the momentum of an object is given by :

p = mv

Where

m is the mass of the duck

[tex]m=\dfrac{p}{v}\\\\m=\dfrac{12\ kg-m/s}{9.1\ m/s}\\\\m=1.31 kg[/tex]

So, the mass of the duck is equal to 1.31 kg.

Answer:

Force,friction,inertia and momentum

Explanation:

The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

Answer:

C) Mass of the ball

Explanation:

Independent variable is the variable the researcher changes.

Since Martin is testing the mass of the ball, he'll be using different balls and that is the only thing he changes.

The distance traveled by the ball is the dependent variable since it depends on the mass of the ball.

The height and length of the ramp are the constant variables since that's the only ones that remain the same throughout this experiment.

Answer:

the induced electric field is 9.95 V/m

Explanation:

Given the data in the question;

Number of turns N = 52

Diameter of coil D = 17 cm = 0.17 m

Radius r = D/2 = 0.17/2 = 0.085 m

Now,

cross-section area A of the coil  = πr²

A = π × ( 0.085 m )²

A = 0.0227 m²

Also given that;

Initial magnetic field B₁ = 0.45 T

Final magnetic field B₂ = 0

∴ change in magnetic field ΔB = B₁ - B₂ = 0.45 T - 0 = 0.45 T

Time taken dT = 0.10 seconds

Now, we know that;

Induced emf ∈ = N[tex](\frac{d\eta }{dt} )[/tex]

where η = BAcosθ

We know that, magnetic field is cylindrically symmetric, coil is also perpendicular to magnetic field.

Hence, the angle between B & A is 0°

∴ θ = 0°

Induced emf ε = N[tex](\frac{d }{dt} )BAcos\theta[/tex]

we substitute

ε = N[tex](\frac{d }{dt} )[/tex] (BAcos0°)

A is constant and cos0° = 1

so

ε = NA[tex](\frac{dB }{dt} )[/tex]  

We now substitute in our values;

ε = 52 × 0.0227 m² ×  [tex](\frac{0.45T }{0.10s} )[/tex]

ε  =  5.3118 V

we know that, from the relation between electric and emf

ε = ∫∈.dl or ε = ∈∫dl         { for coil; ∫dl = πD }

so we have;

ε = ∈πD

solve for ∈

∈ = ε/πD

we substitute

∈ = 5.3118 V / ( π × 0.17 m )

∈ = 9.95 V/m

Therefore, the induced electric field is 9.95 V/m

Answer:

option c

Heat is transferred to your hand from a warm cup

conduction is the process of transferring of heat from one material to another when they are in contact

hope it helps