your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight? ​

Answer:

Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2

Explanation:

Volume of Circular cone = V = (1/3)πr2h

where r is the radius in meters

and h is the height in meters

Substituting the given values in above equation, we get -

V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic  meters.

The force required will be equal to the mass of water in the cone

[tex]= 167.47 * 1000[/tex]

= 167470 Kg

Weight = Mass * g

= 167470 * 10

= 1674700 Kgm/s^2

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Answer:

Explanation:

Given that,

The current in the light bulb, I = 0.5 A

(a) We know that,

Electric current = charge/time

or

Q = It

Put t = 2 hours = 7200 s

So,

Q = 0.5 × 7200

Q = 3600 C

(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]

Let there are n electrons flow through the bulb in 2 hours.

I = Q/t

Since, Q = ne

So,

I = ne/t

[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]

Hence, this is the required solution.

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:It will take about 3000 years

Explanation:

Answer:

a)  k = 701.8 N / m, b)  m_{ast} = 61.1 kg, c)  v ’= -1.3 10⁻⁴ m / s

Explanation:

a) For this exercise let's use the relationship of the angular velocity

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

          k = w² m

the angular velocity is related to the period

          w = 2π / T

we substitute

          k = 4 π²    [tex]\frac{m}{T^2}[/tex]

let's calculate

          k = 4 π²   10 /0.75²

          k = 701.8 N / m

b) now repeat the measurement with an astronaut on the chair

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

where the mass Month the mass of the chair plus the mass of the astronaut

        M = m + [tex]m_{ast}[/tex]

          M = k / w²

          w = 2π / T

let's calculate

           w = 2π / 2

            w = π rad / s

            M = 701.8 /π²

            M = 71,111 kg

now we use that

          M = m + m_{ast}

          m_{ast} = M - m

          m_{ast} = 71.111 - 10.0

          m_{ast} = 61.1 kg

c) if the astronaut's movement is simple harmonic

          x = A cos wt

therefore the speed is

         v = [tex]\frac{dx}{dt}[/tex]

         v = -Aw sin wt

maximum speed is

          v = - Aw

          v = 0.100 π

          v = 0.31416 m / s

we can suppose that the movement of the space station and the astronaut  is equivalent to division of the same

initial instant. Before the move

         p₀ = 0

final instant. When the astronaut is moving

        p_f = M_station v’+ m_{ast} v

the moment is preserved

         p₀ = pf

         0 = M__{station} v ’+ m_{ast} v

         v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]

we substitute

         v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]

         v ’= -1.3 10⁻⁴ m / s

the negative sign indicates that the station is moving in the opposite direction from the astronaut

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Answer:

[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

Explanation:

[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]

[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]

m = Mass of each lump = [tex]30\ \text{g}[/tex]

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]

The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].

Answer:

okay I'm a bit confused but I like the little emoji dudw

Answer:

?

Explanation:

.

Answer:

    q = 2 10⁻⁸ C

Explanation:

For this exercise we use the translational equilibrium equation

                    F_e -A =

                    F_e = W

the electric force is given by Coulomb's law

                    F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case they indicate that the loads on the tapes are equal

                    F_e = k q² / r²

we substitute

                    k q² / r² = m g

                    q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]

calculate  

                     q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]    

                     q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]

                     q = 1,999 10⁻⁸ C

                     q = 2 10⁻⁸ C

Answer: 0.435 m

Explanation:

Given

mass m=20 kg

initial speed u=2 m/s

coefficient of kinetic friction [tex]\mu_k=0.3[/tex]

deceleration which opposes the motion is given by

[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]

[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]

using [tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]

Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

Answer:

roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge

Explanation:

Answer:

2.66×10⁻⁹ N.

Explanation:

From the question,

Applying newton's law of universal gravitation,

Fg = GMm/r²............................... Equation 1

Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury

Given: M = 3.31023 kg, M = 70 kg, r = 2.4106

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)

Fg = 2.66×10⁻⁹ N.

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

Answer:

Explanation:

.

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

Answer:

is there a. b.  c  or d?

Explanation:

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

D. A hockey puck sliding across ice

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Answer:

42,250

Explanation:

It goes inside=

Displacemt

It does work=

Work done

To find efficiency of jule we do=

Dicplacement × Work done

650 × 65

42,250

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