PHYSICS COLLEGE

On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s) ?

Answers

Answer 1

Answer:

Answer:

$P=13.5 W$

Explanation:

In this case, power is the rate of transferring heat per unit time:

$P=(Q)/(\Delta t)(1)$

The heat is given by the formula of the latent heat of fusion, since the ice is melting.

$Q=mL_f(2)$

Here m is the ice's mass and $L_f$ is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

$P=(mL_f)/(\Delta t)\nP=(3.5kg(334*10^3(J)/(kg)))/(86400s)\nP=13.5 W$

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = (1)/(2)mv^(2) + 0$

$2gH = v^(2)$

$v = √(2* 9.8* 5) = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0* 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = (1)/(2)gt^(2)$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8) = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56* 0.638 = 4.823\ m$

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= (1)/(2)mv^2$

$v = √(2gh)$

$v = √(2* 9.8 * 5)$

v = 9.89 m/s

now he perform elastic collision

$v = (m_1v_1)/(m_1+m_2)$

$v = (65* 9.89)/(65+20)$

v = 7.57 m/s

time taken by the skies to fall is

$h = (1)/(2)gt^2$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8)}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Answers

Answer:

a) $f=368.025\ \textup{Hz}$

b) $f_2=736.051\ \textup{Hz}$

Explanation:

Given:

The distance between two speakers (d) = 0.932 m

The distance of the microphone from the midpoint = 2.83 m

Thus, distance of microphone from the nearest speaker (L) = 2.83 - (0.932/2) = 2.364 m

also, the distance of the microphone from the farther speaker (L') = 2.83 + (0.932/2) = 3.296 m

Now,

The path difference is calculated as

L' - L = d = 0.932 m

Now,for a maxima to be produced at the microphone, the waves must constructively interfere.

for this to happen the path difference should be integral multiple of the wavelength.

thus,

$\textup d = n\lambda$

hence, the largest wavelength will be for n = 1,

therefore,

0.932 = 1 × λ

λ = 0.932 m

now, the velocity of sound is given as c = 343 m/s

thus, the frequency will be

$f=(c)/(\lambda)$

on substituting the values, we get

$f=(343)/(0.932)=368.025\ \textup{Hz}$

now, the 2nd largest wavelength will be for n = 2

0.932 = 2 × λ

λ = 0.466

thus, the frequency will be

$f_2=(343)/(0.466)=736.051\ \textup{Hz}$

hence, these are the lowest first two frequencies.

Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answers

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

$v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }$

$v=\sqrt{ 638.6^(2) +49 ^(2) }$

V= 640.48 m/s : total velocity in t= 5s

2. $v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s$

$v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s$

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna

Answers

Answer:

-0.00152 V

Explanation:

Parameters given:

Diameter of the loop = 11 cm = 0.11m

Rate of change of magnetic field, dB/dt = 0.16 T/s

Radius of the loop = 0.055m

The area of the loop will be:

A = pi * r²

A = 3.142 * 0.055²

A = 0.0095 m²

The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:

EMF = - N * A * dB/dt

In this case, there's only one loop, so N = 1.

Therefore:

EMF = -1 * 0.0095 * 0.16

EMF = -0.00152 V

The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.

Answer:

The induced emf is 0.00152 V

Explanation:

Given data:

d = 11 cm = 0.11 m

$r=(d)/(2) =(0.11)/(2) =0.055m$

The area is:

$A=\pi r^(2) =\pi *(0.055^(2) )=0.0095m^(2)$

The induced emf is:

$E=-A(dB)/(dt) =-(0.0095)*(0.16)=-0.00152V$

The negative indicates the direction of E.

An underwater sound source emits waves of frequency 30 kHz in all directions. How does the intensity of the waves (in Watts/m2) vary with distance r from the source?a) 1/r^3
b) 1/r^2
c) 1/r
d) None of above

Answers

When an underwater sound source emits waves of frequency 30 kHz in all directions, the intensity of the waves (in Watts/m2) vary with distance r from the source by the relation 1/r²

As the intensity mechanical sound wave is inversely proportional to the square of the distance from the source, therefore the correct option is B.

What is the Wavelength?

Wavelength can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The intensity of a mechanical wave is inversely proportional to the square of the distance from the source.

An underwater sound source emits waves of frequency of 30 kHz in all directions, the intensity of the waves (in Watts/m2) varies with distance r from the source by the relation 1/r², therefore the correct option is B.

Learn more about wavelength from here

brainly.com/question/7143261

#SPJ2

b or c not sure but try

22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

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