A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off

Complete question is;

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.

Answer:

Tangential speed of foot just before it lands is; v = 5.37m/s

Explanation:

Let U (potential energy) be zero on the ground.

So, initially, U = mgh

where, h = 0.98/2 = 0.49m (midpoint of the leg)

Now just before the leg hits the floor it would have kinetic energy as;

K = ½Iω²

where ω = v/r and I = ⅓mr²

So, K = ½(⅓mr²)(v/r)²

K = (1/6) × (mr²)/(v²/r²)

K = (1/6) × mv²

From principle of conservation of energy, we have;

Potential energy = Kinetic energy

Thus;

mgh = (1/6) × mv²

m will cancel out to give;

gh = (1/6)v²

Making v the subject, we have;

v = √6gh

v = √(6 × 9.81 × 0.49)

v = √28.8414

v = 5.37m/s

Answer:

A.) 78.4 J for both

B.) 78.4 J for both

C.) 8.85 m/s for both

D.) 17.7 kgm/s

Explanation:

Given information:

Mass m = 2 kg

Distance d = 20 m

High h = 4 m

A.) Gravitational potential energy can be calculated by using the formula

P.E = mgh

P.E = 2 × 9.8 × 4

P.E = 78.4 J

Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J

B.) According to conservative energy,

Maximum P.E = Maximum K.E.

Therefore, the kinetic energy of the two blocks will be 78.4 J

C.) Since K.E = 1/2mv^2 = mgh

V = √(2gh)

Solve for velocity V by substituting g and h into the formula

V = √(2 × 9.8 × 4)

V = √78.4

V = 8.85 m/s

The velocities of both block will be 8.85 m/s

D.) Momentum is the product of mass and velocity. That is,

Momentum = MV

Substitute for m and V into the formula

Momentum = 2 × 8.85 = 17.7 kgm/s

Both block will have the same value since the ramp Is frictionless.

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:

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

Explanation:

(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:

[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]

(b) The frequency of a wave is inversely proportional to its period:

[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]

(c) The wavelength is the distance between two successive crests, so:

[tex]\lambda=30.2m[/tex]

(d) The speed of a wave is defined as:

[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]

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:

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

Explanation:.

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:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

[tex]a=\sqrt{a_c^2+a_T^2}[/tex]         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

[tex]a_c=r\omega^2[/tex]

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]        

Next, you calculate the tangential acceleration. aT is calculate by using:

[tex]a_T=r\alpha[/tex]    (2)

α: angular acceleration

The angular acceleration is:

[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]

The total acceleration of the ball is 14.022 m/s²

Answer:

10.15m/s

Explanation:

The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:

f₁ = [(v ± v₁) / (v ± v₂)] f            ----------------------(i)

Where;

f₁ = frequency received by the observer or receiver

v = speed of sound in air

v₁ = velocity of the observer

v₂ = velocity of the source

f = original frequency of the sound

From the question, the observer is the bicyclist and the source is the car driver. Therefore;

f₁ = frequency received by the observer (bicyclist) = 357Hz

v = speed of sound in air = 330m/s

v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂

v₂ = velocity of the source (driver)

f = original frequency of the sound = 365Hz

Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.

From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.

i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.

Substitute these values into equation (i) as follows;

357 = [(330 - 0.33v₂) / (330 - v₂)] * 365

(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]

0.98 =  [(330 - 0.33v₂) / (330 - v₂)]

0.98 (330 - v₂) =  (330 - 0.33v₂)

323.4 - 0.98v₂ = 330 - 0.33v₂

323.4 - 330 = (0.98 - 0.33)v₂

-6.6 = 0.65v₂

v₂ = -10.15

The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.

iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.

Substitute these values into equation (i) as follows;

357 = [(330 + 0.33v₂) / (330 + v₂)] * 365

(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]

0.98 =  [(330 + 0.33v₂) / (330 + v₂)]

0.98 (330 + v₂) =  (330 + 0.33v₂)

323.4 + 0.98v₂ = 330 + 0.33v₂

323.4 - 330 = (0.33 - 0.98)v₂

-6.6 = -0.65v₂

v₂ = 10.15

The value of v₂ is positive and that is a valid solution.

Therefore, the speed of the car is 10.15m/s

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

[tex]W_c=Mgh[/tex]        (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex]     (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

[tex]0.16(E_c)=4,164,294.4J[/tex]

Ec: Calories

You solve for Ec and convert the result to Cal:

[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]

The amount of Calories needed by the climber was 6220.56 kcal

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 maximum amount of meat that the butcher can sell is  [tex]3\frac{1}{12}\:lb[/tex]

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]

Best Regards!

Answer:

0 - Then, the ray is totally reflected

Explanation:

The ray reaches the boundary between the two mediums at 49.2°.

If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.

You use the following to calculate the critical angle:

[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex]       (1)

n2: index of refraction of the second medium (air) = 1.00

n1: index of refraction of the first medium (glass) = 1.51

You replace the values of the parameters in the equation (1):

[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]

The critical angle is 41.47°, which is lower than the incident angle 49.2°.

Then, the ray is totally reflected.

0

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:

(a) T = 1.35 s

(b) vmax = 0.17 m/s

(c) v = 0.056 m/s

Explanation:

(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:

[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]          (1)

m: mass of the cart = 350 g = 0.350kg

k: spring constant = 7.5 N/m

[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]

The period of oscillation of the car is 1.35s

(b) The maximum speed of the car is given by the following formula:

[tex]v_{max}=\omega A[/tex]       (2)

w: angular frequency

A: amplitude of the motion = 3.8 cm = 0.038m

You calculate the angular frequency:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]              

Then, you use the result of w in the equation (2):

[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]

The maximum speed if 0.17m/s

(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:

[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]

The speed is the value of the following function for t = 0.069s

[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]

The speed of the car is 0.056m/s

Answer:

a) 8.33 x [tex]10^{-6}[/tex] Pa  or  8.22 x [tex]10^{-11}[/tex] atm

b) 1.66 x [tex]10^{-5}[/tex] Pa  or  1.63 x [tex]10^{-10}[/tex] atm

c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Explanation:

Intensity of light = 2500 W/m^2

area = 25 ft^2

a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]

where I is the intensity of the light

c is the speed of light = [tex]3*10^{8} m/s[/tex]

[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa

1 pa = [tex]9.87*10^{-6}[/tex]

8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm

b) average radiation for a totally radiating section of the floor

[tex]Pav = \frac{2I}{c}[/tex]

this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section

therefore,

Pav = 2 x 8.33 x [tex]10^{-6}[/tex]  = 1.66 x [tex]10^{-5}[/tex] Pa

or

Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm

c) average momentum per unit volume is

[tex]m = \frac{I}{c^{2} }[/tex]

[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Answer:

Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.

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:

[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:

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:

% increase in resistance of tungsten = 9.27%

Explanation:

We are given:

Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C

Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C

% Resistance change of gold wire due to temperature change = 7%

Now, let R1 and R2 be the resistance before and after the temperature change respectively.

Thus;

(R2 - R1)/R1) x 100 = 7

So,

(R2 - R1) = 0.07R1

R2 = R1 + 0.07R1

R2 = 1.07R1

The equation to get the change in temperature is given as;

R2 = R1(1 + αΔt)

So, for gold,

1.07R1 = R1(1 + 0.0034*Δt)

R1 will cancel out to give;

1.07 = 1 + 0.0034Δt

(1.07 - 1)/0.0034 = Δt

Δt = 20.59°C

For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;

Rt2 = Rt1(1 + α_t*Δt)

So, Rt2/Rt1 = 1 + 0.0045*20.59

Rt2/Rt1 = 1.0927

From earlier, we saw that;

(R2 - R1)/R1) x 100 = change in resistance

Similarly,

(Rt2 - Rt1)/Rt1) x 100 = change in resistance

Simplifying it, we have;

[(Rt2/Rt1) - 1] × 100 = %change in resistance

Plugging in the value of 1.0927 for Rt2/Rt1, we have;

(1.0927 - 1) × 100 = %change in resistance

%change in resistance = 9.27%

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;

The pond's water level will fall.

Explanation;

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Hope this Helps!!!

Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it

The pond water level will lower slightly

According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced

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Answer: not sure

Explanation:

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:

The value of the potential energy of the particle at point B is 85 joules.

Explanation:

According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:

[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]

Where:

[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.

[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.

[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.

[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.

The initial kinetic energy of the particle is:

[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Velocity, measured in meters per second.

If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:

[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]

[tex]K_{A} = 20\,J[/tex]

Finally, the value of the potential energy at point B is:

[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]

[tex]U_{B} = 85\,J[/tex]

The value of the potential energy of the particle at point B is 85 joules.

The potential energy of the particle at point B is 85 J.

Given to us:

Mass of the particle, [tex]m=0.40\ kg[/tex]

velocity of the particle, [tex]v= 10\ m/s[/tex]

potential energy of the particle, [tex]PE= 40\ J[/tex]

Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]

Calculating the kinetic energy of the particle,

[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]

According to the  Principle of Energy Conservation,

The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,

Also,

Total Energy at point A ,

[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]

Total Energy at point B,

[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]

As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,

[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]

Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.

Hence, the potential energy of the particle at point B is 85 J.

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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