A disk with a rotational inertia of 2.5 kg-m2 and a radius 1.1 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 7.7 N is applied tangentially to the rim. The angular acceleration of the disk is _____ rad/s2. Round your answer to the nearest tenth.

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

Answer:

The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]

So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Answer:

Option A, Sea Breeze

Explanation:

Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.

Hence, option A is correct

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]

[tex]kx^2 = mv^2[/tex]

[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]

[tex]7.2 =200*10^{-3}*v^{2}[/tex]

[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]

[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

Answer:

Check the explanation

Explanation:

Part A

F = CA

this drag force balances the weight = 6X 9.8

so

6X9.8 = 0.5 X A X0.5 X 1.2 X 532

A= 0.069 m2

Part B

here the sorce is moving and the observer is at rest

so f= f(- 1 - 1

f = 1.1X10 343 343 – 53

f' = 1.3 KHz

Part C:

given the intensity = 30 dB

we know that I dB = 10 log (I(W/m2))

so we get I (W/m2) = 1000

Part D : The catch

Given that U1 = 53 M1 = 6 kg

U2 =-10 M2=0.25

V1=V2

now conserving momentum

6 X 53 -0.25 X10 =(6+0.25)V

V= 50.48 m/sec

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

Answer: R.A.M

Explanation:

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

Answer:147 joules

Explanation:

Mass=m=5kg

Acceleration due to gravity=g=9.8m/s^2

Height=h=3 meter

Potential energy=m x g x h

Potential energy=5 x 9.8 x 3

Potential energy=147 joules

Answer:

The thickness of the door is 0.4230 m

Explanation:

Given;

mass of bullet, m = 0.009 kg

initial velocity of the bullet, u = 803 m/s

final velocity of the bullet, v = 617 m/s

average resistive force of the door on the bullet, F = 5620 N

Apply Newton's second law of motion;

Force exerted by the door on the bullet = Force of the moving bullet

F = ma

where;

F is applied force

m is mass

a is acceleration

Also, Force exerted by the door on the bullet = Force of the moving bullet

[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]

where;

v is the final velocity of the bullet

u is initial velocity of the bullet

t is time

We need to calculate the time spent by the bullet before it passes through the door.

[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]

Distance traveled by the bullet within this time period = thickness of the door

This distance is equivalent to the product of average velocity and time

[tex]S = (\frac{u+v}{2}) t[/tex]

where;

s is the distance traveled

[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]

Therefore, the thickness of the door is 0.4230 m

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Answer:

The answer is A

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent

Answer:

A.

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent.

Answer:

Mass

Explanation:

Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.  

Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.  

Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2

Answer:

The Object will immediately begin moving toward the left

Explanation:

Because the force of thirteen is greater than ten and applied to the opposite side

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

Answer:

Mercury is rocky

Explanation:

Answer:

Rocky

Explanation:

It has no atmosphere so it cannot hold gas.

Answer:

L= 276.4 mm

Explanation:

Given that

E= 180 GPa

d= 3.7 mm

F= 1890 N

ΔL= 0.45 mm

We know that ,elongation due to load F in a cylindrical bar is given as follows

[tex]\Delta L =\dfrac{FL}{AE}[/tex]

[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]

Now by putting the values in the above equation we get

[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]

L=0.2764 m

L= 276.4 mm

Therefore the length of the specimen will be 276.4 mm

Answer:

Explanation:

The problem is based on Newton's law of cooling .

According to Newton's law

dQ / dt = k ( T - T₀ ) ,

dT / dt = k' ( T - T₀ )          ; dT / dt is rate of fall of temperature.

T is average  temperature of hot body , T₀ is temperature of surrounding .

In the first case rate of fall of temperature = (210 - 191) / 2.5

= 7.6 degree / s

average temperature T = (210 + 191) /2

= 200.5  

Putting in the equation

7.6 = k' ( 200.5  - 64 )

k' = 7.6 / 136.5

= .055677

In the second case :---

In the second case, rate of fall of temperature = (191 - 156) / t  

= 35 / t   , t is time required.

average temperature T = (156 + 191) /2

= 173.5  

Putting in the equation

35 / t = .05567 ( 173.5 - 64 )

t = 5.74 minute .

Answer:

Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.

Explanation: Hope i helped!!!

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

Explanation:

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

There will be no change in the angular momentum of the system.

Explanation:

Total angular momentum of the system  will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .  

Answer:

Here's your answer :

hope it helps!

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]

Next, you use the formula for the magnetic force produced by the wires:

[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Answer:

The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

Explanation:

Information from the question:

Number of turns of the coil = 100 turns

The diameter of the coil = 6 m

The radius of the coil = diameter / 2 = 3 m

The coil current = 2.5 A

Formula : The Magnetic field at the center of the coil =

                                  k * number of turns * current / 2 * radius

Therefore, The Magnetic field at the center of the coil=

                                 (4 * [tex]\pi[/tex] * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)

The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

Answer:

Answer: the true statement form the given statements is “the athletes is not doing any work because he does not move weight”

Explanation:

The athlete isn’t doing any work because he doesn’t move the weight is the correct statement.

Here ,

Hence there is no work done by the athlete

Therefore ,The athlete isn’t doing any work because he doesn’t move the weight is the correct statement

Learn more about Work done here

https://brainly.com/question/25573309

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Answer:

The current pass the [tex]R_2[/tex] is  [tex]I = 0.25 A[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  [tex]V = 3V[/tex]

     The first resistance is  [tex]R_1 = 7 \Omega[/tex]

     The second resistance is  [tex]R_2 = 5 \Omega[/tex]

Since the resistors are connected in series their equivalent resistance is  

       [tex]R_{eq} = R_1 +R_2[/tex]

Substituting values

         [tex]R_{eq} = 7 + 5[/tex]

         [tex]R_{eq} = 12 \Omega[/tex]

Since the resistance are connected in serie the current passing through the circuit  is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as

        [tex]I = \frac{V}{R_{eq}}[/tex]

Substituting values  

      [tex]I = \frac{3}{12}[/tex]

      [tex]I = 0.25 A[/tex]