An object of mass 3.07 kg, moving with an initial velocity of 5.07 m/s, collides with and sticks to an object of mass 2.52 kg with an initial velocity of -3.11 m/s. Find the final velocity of the composite objec

Answer:

[tex]v_B=3.78\times 10^5\ m/s[/tex]

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]

Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]

Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]

Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]

So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

Answer:

I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right

Explanation:.

Complete question:

Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.

Answer:

The ratio of the power delivered to A to power delivered to B is 7 : 1

Explanation:

Cross sectional area of a wire is calculated as;

[tex]A = \frac{\pi d^2}{4}[/tex]

Resistance of a wire is calculated as;

[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]

Resistance in wire A;

[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]

Resistance in wire B;

[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]

Power delivered in wire;

[tex]P = \frac{V^2}{R}[/tex]

Power delivered in wire A;

[tex]P = \frac{V^2_A}{R_A}[/tex]

Power delivered in wire B;

[tex]P = \frac{V^2_B}{R_B}[/tex]

Substitute in the value of R in Power delivered in wire A;

[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]

Substitute in the value of R in Power delivered in wire B;

[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]

Take the ratio of power delivered to A to power delivered to B;

[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]

The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]

The wires are connected across the same potential; [tex]V_A = V_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]

wire A has seven times the diameter and seven times the length of wire B;

[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]

Therefore, the ratio of the power delivered to A to power delivered to B is

7 : 1

Answer:

46648 J

Explanation:

mass m= 85 Kg

velocity v = 56 m/s

distance covered s =40 m

According to Question,

frictional drag force to him that matched his weight

[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]

Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J

W=Q= F_d×s

=833×56 = 46648 J

Answer:

17.656 m

Explanation:

Initial speed u = 20 m/s

angle of projection α = 60°

at the top of the trajectory, one fragment has a speed of zero and drops to the ground.

we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.

To find the range of the projectile, we use the equation

R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]

where g = acceleration due to gravity = 9.81 m/s^2

Sin 2α = 2 x (sin α) x (cos α)

when α = 60°,

Sin 2α  = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5

Sin 2α  = 0.866

therefore,

R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m

since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m

Answer:

a) Jack does more work uphill

b) Numerically, we can see that Jill applied the most power downhill

Explanation:

Jack's mass = 75 kg

Jill's mass = [tex]1.5x = 75[/tex]

Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg

distance up hill = 15 m

a) work done by Jack uphill = mgh

where g = acceleration due to gravity= 9.81 m/s^2

work = 75 x 9.81 x 15 = 11036.25 J

similarly,

Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J

this shows that Jack does more work climbing up the hill

b) assuming Jack's time downhill to be t,

then Jill's time = [tex]\frac{t}{3}[/tex]

we recall that power is the rate in which work id done, i.e

P = [tex]\frac{work}{time}[/tex]

For Jack, power = [tex]\frac{11036.25}{t}[/tex]

For Jill, power =  [tex]\frac{3*7357.5}{t}[/tex] =  [tex]\frac{22072.5}{t}[/tex]

Numerically, we can see that Jill applied the most power downhill

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.

Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]

The orbital speed for a satellite is given by the formula as follows :

[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]

So, the orbital speed for a satellite is 349.23 m/s.

Answer:

F = 103.54N

Explanation:

In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.

The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.

Furthermore, you have that the sum of forces are given by:

[tex]F-Wsin\theta=0[/tex]                (1)

F: applied force = ?

W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N

θ: degree of the incline = 25°

You solve the equation (1) for F:

[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex]          (2)

The applied force on the box is 103.54N

Answer:

88.29 N

Explanation:

mass of the ball = 4.5 kg

weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)

weight W = 4.5 x 9.81 = 44.145 N

centrifugal forces Tc act on the ball as it swings.

At the top point of the vertical swing,

Tension on the rope = Tc - W.

At the bottom point of the vertical swing,

Tension on the rope = Tc + W

therefore,

difference in tension between these two points will be;

Net tension = tension at bottom minus tension at the top

= Tc + W - (Tc - W) = Tc + W -Tc + W

= 2W

imputing the value of the weight W, we have

2W = 2 x 44.145 = 88.29 N

Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]

The formula for the velocity of the cell is

[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]

[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]

[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]

The maximum acceleration of the cells as they move toward each other and just barely touch is

[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]

[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]

[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]

The answers obtained are;

1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].

2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].

The answer is explained as shown below.

We have, the mass of the red blood cell;

Also, the charges of the cells are;

The distance between the charges when they barely touch will be two times the radius of each charge.

1. As both the cells are negatively charged they will repel each other.

Now, substituting all the known values in the equation, we get;

[tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

Substituting the known values, we get;

[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]

[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]

Find out more information about moving charges here:

https://brainly.com/question/14632877

Answer:

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]

hope this helps ..

Good luck on your assignment..

Answer:

The astronauts are separated by 28 m.

Explanation:

The separation of the astronauts can be found by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}v_{1f} = -m_{2}v_{2f} [/tex]

[tex] v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60} [/tex]

Now, the distance (x) is:      

[tex] x = \frac{v}{t} [/tex]  

The distance traveled by the astronaut 1 is:

[tex] x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t [/tex]    (1)

And, the distance traveled by the astronaut 2 is:

[tex] x_{2} = v_{2f}*t [/tex]  (2)

From the above equation we have:  

[tex] t = \frac{x_{2}}{v_{2f}} [/tex]    (3)                                    

By entering equation (3) into (1) we have:    

[tex] x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}}) [/tex]

[tex] x_{1} = -\frac{4*12}{3} = -16 m [/tex]    

The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.      

Finally, the separation of the astronauts is:

[tex] x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m [/tex]

Therefore, the astronauts are separated by 28 m.

I hope it helps you!

The total separation between the two astronauts is 28m.

The given parameters:

Apply the principle of conservation of momentum to determine the final velocity of each astronauts as follows;

[tex]m_1u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\60(0) + 80(0) = 60(v_1) + 80(v_2)\\\\0 = 60v_1 + 80v_2\\\\-60v_1 = 80v_2\\\\v_1 = \frac{-80v_2}{60} \\\\v_1 = -1.333v_2[/tex]

Let the time when astronaut 2 moved 12 m = t

The distance traveled by astronaut 1 is calculated as;

[tex]x_1 = v_1 t\\\\x_1 = -1.333v_2t[/tex]

The  distance traveled by astronaut 2 is calculated as;

[tex]x_2 = v_2 t\\\\12 = v_2t\\\\t = \frac{12}{v_2}[/tex]

Now solve for the distance of astronaut 1

[tex]x_1 = - 1.333v_2 \times t\\\\x_1 = -1.333 v_2 \times \frac{12}{v_2} \\\\x_1 = -16 \ m[/tex]

The total separation between the two astronauts is calculated as follows;

[tex]d = |x_1| + x_2\\\\d = 16 + 12\\\\d = 28 \ m[/tex]

Learn more about conservation of linear momentum here: https://brainly.com/question/24424291

Answer:

Umax = 105.8nJ

Umin =-105.8nJ

Umax-Umin = 211.6nJ

Explanation:

Answer:

According to Newtons 2nd law of motion ;

  The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

This law is simply saying ;

Force = Mass ×Acceleration

I Hope It Helps  :)

Answer:

The inner diameter is 57.3 cm

Explanation:

The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):

[tex] W_{s} = W_{w} [/tex]

[tex] m_{s}*g = m_{w}*g [/tex]            

[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]    

Where D is the density and V is the volume

[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]    

Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter    

[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]                    

[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]          

[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]  

[tex] d_{i} = 57.3 cm [/tex]                  

Therefore, the inner diameter is 57.3 cm.    

I hope it helps you!    

Answer:

The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Explanation:

to solve this, we will have to apply the knowledge that will be got from the equations of motion.

There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.

In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula

[tex]v = u +at[/tex]

where v= final velocity = 0

u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]

t= time = 4.29 seconds.

[tex]a = \frac{v - u}{t}[/tex]

[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]

Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Explanation :

Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

The beta decay reaction is:

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Answer:

the blank is 39

Explanation: a p e x

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        [tex]Em_{f}[/tex] = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     [tex]Em_{f}[/tex] = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

Answer:

[tex]\theta=65.18rad[/tex]

Explanation:

The angle in rotational motion is given by:

[tex]\theta=\frac{w_o+w_f}{2}t[/tex]

Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:

[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]

The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:

[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]

Answer:

The temperature is  [tex]T = 168.44 \ K[/tex]

Explanation:

From the question ewe are told that

   The rate of heat transferred is    [tex]P = 13.1 \ W[/tex]

     The surface area is  [tex]A = 1.55 \ m^2[/tex]

      The emissivity of its surface is  [tex]e = 0.287[/tex]

Generally, the rate of heat transfer is mathematically represented as

           [tex]H = A e \sigma T^{4}[/tex]

=>         [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]

where  [tex]\sigma[/tex] is the Boltzmann constant with value  [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]

substituting value  

             [tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]

            [tex]T = 168.44 \ K[/tex]

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

We know that the density of the ice is smaller than the density of the water (and this is why the ice floats in water).

Dw > Di

Da is the density of the water and Di is the density of the ice

Since in Bowl A we have a volume V, only of water, then the mass of the bowl A is:

Dw*V.

Now, in the bowl B we have a combination of water and ice, suppose that Vw is the volume of water and Vi is the volume of ice, and we know that:

Vw + Vi = V.

Then the mass in this second bowl is:

Dw*Vw + Di*Vi = Dw*(V - Vi) + Di*Vi = Dw*V + (Di - Dw)*Vi

and we know that Dw > Di, then the left term is a negative term, then the mass of bowl B is smaller than the mass of bowl A.

Answer:

Q = 5.9 nC (Approx)

Explanation:

Given:

Further distance = 10 cm

Electric potential(V) = 190 v

Potential difference(V1) = 140 v

Find:

Magnitude of the charge Q

Computation:

V = KQ / r

190 = KQ / r.............Eq1

V1  = KQ / (r+10)

140 = KQ / (r+10) ............Eq2

From Eq2 and Eq1

r = 28 cm = 0.28 m

So,

190 = KQ / r

190 = (9×10⁹)(Q) / 0.28

53.2 = (9×10⁹)(Q)

5.9111 = (10⁹)(Q)

Q = 5.9 nC (Approx)

Answer:

d) Both blocks have had equal losses of energy to friction

Explanation:

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces

Moreover, the block A is twice as fast than block B at the time of crossing the finish line

So based on the above information,  it contains the losses of identical friction

And we also know that

Friction energy loss is

[tex]= \mu \times m \times g \times D[/tex]

It would be the same for both the blocks

hence, the option d is correct

The correct answer will be both blocks have had equal losses of energy to friction.

Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.

Moreover, the block A is twice as fast as block B at the time of crossing the finish line.

So based on the above information,  it contains the losses of identical friction.

And we also know that

Friction energy loss is

[tex]E_f=\mu m g D[/tex]

It would be the same for both the blocks

Hence both blocks have had equal losses of energy to friction.

To know more about friction, follow

https://brainly.com/question/24386803

Answer:

[tex]d_2=3.16cm[/tex]

Explanation:

So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.

In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:

[tex]\sum{F_{x}}=0[/tex]

so:

[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]

Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:

[tex]-F_{12}+2F_{13x}=0[/tex]

we can now solve this for [tex]F_{12}[/tex] so we get:

[tex]F_{12}=2F_{13x}[/tex]

Now we can substitute with the electrostatic force formula, so we get:

[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]

We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]

so the simplified equation is:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]

From the given diagram we know that:

[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]

so when solving for [tex]r_{13}[/tex] we get:

[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]

and if we square both sides of the equation, we get:

[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]

and we can substitute this into our equation:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]

so we can now solve this for [tex]r_{12}[/tex] so we get:

[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

which can be rewritten as:

[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

and now we can substitute values.

[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]

which solves to:

[tex]r_{12}=6.16cm[/tex]

now, we must find [tex]d_{2}[/tex] by using the following equation:

[tex]r_{12}=d_{1}+d_{2}[/tex]

when solving for [tex]d_{2}[/tex] we get:

[tex]d_{2}=r_{12}-d_{1}[/tex]

when substituting we get:

[tex]d_{2}=6.16cm-3cm[/tex]

so:

[tex]d_{2}=3.16cm[/tex]

Answer: -254.51°O

Explanation:

Ok, in our scale, we have:

-182.9°C corresponds to 100° O

-218.3°C corresponds to 0°

Then we can find the slope of this relation as:

S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C

So we can have the linear relationship between the scales is:

Y = (2.82°O/°C)*X + B

in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.

And we know that when X = -182.9°C, we must have Y = 0°O

then:

0 = (2.82°O/°C)*(-182.9°C) + B

B = ( 2.82°O/°C*189.9°C) = 515.778°O.

now, we want to find the 0 K in this scale, and we know that:

0 K = -273.15°C

So we can use X =  -273.15°C in our previous equation and get:

Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O