Two objects which have a total mass of 12.0 kg are placed adjacent to one another. The object on the left is 7.4 kg, and it pushed to the right with a force of 37 Newtons. With what force does the left object push against the right object?

Answers

Answer 1

Answer:

22.89 N

Explanation:

F = 37 N

Let the acceleration in the system is a and f be the force between the tewo blocks.

Apply the Newton's second law

By the free body diagrams

F - f = 7.4 x a .... (1)

f = 12 x a ...... (2)

Adding both of them

37 = 19.4 x a

a = 1.9 m/s^2

Put in equation (2)

f = 12 x 1.9 = 22.89 N

Two Objects Which Have A Total Mass Of 12.0 Kg Are Placed Adjacent To One Another. The Object On The

Related Questions

When the particle with charge q reaches the center of the original square, it is, as stated in the problem, momentarily at rest. Is the particle at equilibrium at that moment?

Answers

Answer:

NO

Explanation:

The answer is NO because when the particle reaches the center of square there is a net force acting on the particle dew to various other charges and this net force gives acceleration to the particle. Moreover, For particle or object to be in equilibrium the net force acting on it should be zero and hence no acceleration. Although velocity can be zero or non zero at equilibrium state.

Answer:

The context is missing here, but ill try to explain a general case.

Something is in equilibrium if it is in a valley of the potential energy, this is because things in life try to be in the minimal energy state possible. Think for example in a thing that is away from the ground, the object will try to reach the ground, in this way minimizing the potential energy.

Now, if once the particle reaches the center of the square it remains at rest, it means that the total forces acting on the particle are zero and this is why the particle stays at rest, this would mean that the particle is in equilibrium, and if someone moves it a little bit of the center, some of the forces will increase and others will decrease, and then the equilibrium will be broken and the particle will move again.

In another case, if the particle is momentarily at rest (just for a few seconds) it may be because the forces acting on it are affecting the particle in such way that is moving is fully stopped in one direction, and the new forces are accelerating the particle in the opposite direction (in the same way that if you throw something upside when it reaches the maximum height it has for a brief moment a velocity equal to zero)  

A jet airplane with a 75.0 m wingspan is flying at 255 m/s. What emf is induced between the wing tips in V if the vertical component of the Earth's magnetic field is 3.00 x 10^-5 T?

Answers

Answer:

The emf is 0.574 volt.

Explanation:

Given that,

Magnetic field [tex]B =3.00\times10^{-5}\ T[/tex]

Length = 75.0 m

Velocity = 255 m/s

We need to calculate the emf

Using formula of emf

[tex]\epsilon=Blv[/tex]

Where, v = velocity

l = length

B = magnetic field

Put the value into the formula

[tex]\epsilon= 3.00\times10^{-5}\times75.0\times255[/tex]

[tex]\epsilon=0.574\ volt[/tex]

Hence, The emf is 0.574 volt.

Final answer:

The emf induced between the wingtips of a jet airplane flying at 255 m/s with a 75.0 m wingspan in Earth's magnetic field of strength 3.00 x 10^-5 T is 0.57 V. This is calculated using the formula for motional emf: E = Blv.

Explanation:

To calculate the emf induced between the airplane wings due to Earth's magnetic field, we use the formula for motional emf: E = Blv, where B is the magnetic field strength, l is the length of the conductor moving through the field (in this case, the wingspan of the airplane), and v is the velocity of the conductor.

Substituting the given values into the formula we get:
E = (3.00 x 10^-5 T) * (75.0 m) * (255 m/s) = 0.57 V

This means that a 0.57 Volt emf is induced between the wingtips of the airplane due to the Earth's magnetic field. This is a direct application of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an emf in a conductor.

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A solenoid with n1 = 1200 turns/m and a current I1 = 2.5 A is filled with a paramagnetic material at a temperature T1 = 320 K. A second solenoid with n2 = 1000 turns/m and a current I2= 0.85 A is filled with the same paramagnetic material at a different temperature T2. The magnetizations are the same in both cases. What is the value of T2?

Answers

Answer:

[tex]T_2[/tex] = [tex]90.667K[/tex]

Explanation:

Given:

For the first solenoid

Number of turns, n₁ = 1200 turns/m

Current, I₁ = 2.5 A

Paramagnetic material temperature, T₁ = 320 K

Now for the second solenoid

Number of turns, n₂ = 1000 turns/m

Current, I₂ = 0.85 A

Paramagnetic material temperature = T₂

The magnetic flux (B) is given as

[tex]B=\frac{c\mu_onI}{T}[/tex]

where,

c = curie's constant

μ₀ = arithmetic constant

also it is given that the magnetization in both the cases are same

therefore the magnetic flux will also be equal

thus,

[tex]\frac{c\mu_on_1I_1}{T_1}[/tex] = [tex]\frac{c\mu_on_2I_2}{T_2}[/tex]

or

[tex]\frac{n_1I_1}{T_1}[/tex] = [tex]\frac{n_2I_2}{T_2}[/tex]

or

[tex]\frac{1200\times 2.5}{320}[/tex] = [tex]\frac{1000\times 0.85}{T_2}[/tex]

or

[tex]9.375[/tex] = [tex]\frac{850}{T_2}[/tex]

or

[tex]T_2[/tex] = [tex]\frac{850}{9.375}[/tex]

or

[tex]T_2[/tex] = [tex]90.667K[/tex]

Two cars approach an extremely icy four-way perpendicular intersection. Car A travels northward at 10 m/s and car B is traveling eastward. They collide and stick together, traveling at 41.5° north of east. What was the initial velocity of car B (in m/s)? (Enter the magnitude. Assume the masses of the cars are equal.)

Answers

Answer:

v = 11.3 m/s

Explanation:

Since there is no external force on the system of two cars

so here momentum of the system of two cars will remain constant

it is given here that two cars combined and move together at 41.5 degree North of East after collision

so here direction of final momentum is given as

[tex]tan\theta = \frac{P_y}{P_x}[/tex]

now we have initial momentum in the same direction

[tex]P_1 = m(10 m/s)\hat j[/tex]

[tex]P_2 = mv\hat i[/tex]

now we have

[tex]\frac{P_1}{P_2} = tan\theta[/tex]

[tex]\frac{m(10 m/s)}{mv} = tan41.5[/tex]

[tex]\frac{10}{v} = 0.88[/tex]

[tex]v = 11.3 m/s[/tex]

Final answer:

Using the principles of conservation of momentum and trigonometry, the initial velocity of car B can be calculated to be approximately 9.75 m/s.

Explanation:

The problem can be solved using the principles of conservation of momentum. In a two-dimensional collision, the components of the momentum in the X and Y directions are separately conserved. Since we're given that the cars have equal mass and they stick together after colliding to make a single object, the magnitude of the velocity of the combined object after the collision must be equal to the vector sum of their initial velocities.

Car A is moving northward and thus its velocity is contributing only to the Y component of the total momentum. Car B, on the other hand, is moving eastward, so its velocity is contributing to the X component. When the cars collide and stick together, they move at an angle of 41.5° north of east, meaning there is both an X and Y component to their velocity.

Using trigonometry, the X component (eastward direction, i.e., car B's direction) can be found by VB = VA * tan(Θ). Given that VA is 10 m/s and Θ is 41.5 degrees, VB can be calculated to be approximately 9.75 m/s.

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What frequency must a sound wave have in air to have the same wavelength as a 750 Hz sound wave in a platinum bar? Vair = 340 m/s, Vpt = 2800 m/s,

Answers

Answer:

f = 91.1 Hz

Explanation:

As we know that it is given here that wavelength of sound in platinum is same as the wavelength of sound in air

so we can use the formula of wavelength in two mediums

[tex]\frac{speed}{frequency} = \frac{speed}{frequency}[/tex]

now it is given that

[tex]v_{air} = 340 m/s[/tex]

[tex]v_{pt} = 2800 m/s[/tex]

frequency of sound in platinum is 750 Hz

now frequency of sound in air = f

now from above formula

[tex]\frac{340}{f} = \frac{2800}{750}[/tex]

[tex]f = 91.1 Hz[/tex]

A total of 600 C of charge passes through flashlight bulb in 0.500 hr. What is the average current? Give your answer in A.

Answers

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 600C

Δt = 0.500hr = 1800s

Plug in the values and solve for I:

I = 600/1800

I = 0.333A

The average current  flowing in the flashlight bulb is 0.333 A

The given parameters:

quantity of the charge, Q = 600 C

time of the current flow, t = 0.5 hour

To find:

the average current, I

The average current is calculated from quantity of charge that flows in the given time period:

Q = It

where:

I is the average current

Q is the quantity of the charge

t is the time period

From the equation above, make the average current the subject of the formula:

[tex]I = \frac{Q}{t} \\\\I = \frac{600 \ C}{0.5 \ hr \times 3600 \ s} \\\\I = \frac{600 \ C}{1800 \ s} \\\\I = 0.333 \ A[/tex]

Thus, the average current is 0.333 A

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A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. What is the magnitude of the acceleration of the ball when it is at its highest point?

Answers

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

[tex]F_g = mg[/tex]

now the acceleration of ball is given as

[tex]a = \frac{F_g}{m}[/tex]

[tex]a = \frac{mg}{m} = g[/tex]

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Final answer:

The magnitude of the acceleration of a tennis ball at its highest point, in an evacuated chamber and under the influence of gravity alone, is 9.8 m/s². This remains constant regardless of the ball's position in its trajectory.

Explanation:

The question asks about the magnitude of the acceleration of a tennis ball at its highest point when it is shot vertically upward in an evacuated chamber with an initial speed. This scenario is governed by the principles of classical mechanics, specifically under the domain of gravity's influence on objects in motion. Regardless of whether the object is moving up or down or is at its peak height, the acceleration due to gravity on Earth remains constant.

Acceleration at the highest point, or at any point during its motion under gravity alone (assuming no air resistance), is -9.8 m/s2. It's essential to understand that the negative sign indicates the direction of acceleration, which is towards the center of the Earth but does not affect the magnitude of acceleration. Hence, the magnitude of acceleration is 9.8 m/s2, indicating that gravity is the only force acting on the ball throughout its trajectory.

A table-top fan has radius of 0.5 m. It starts to rotate from rest to 800 rpm within 30 seconds. Determine: 1) the angular momentum 2) how many revolutions have it gone through after 2 minutes? 3) If the rotational inertia of each blade around the center is 0.3 kg.m^ 2, what is the magnitude of the torque provided by the motor of the fan, assuming no friction?

Answers

Answer:

Part a)

L = 25.13 kg m^2/s

Part b)

N = 3200 rev

Part c)

torque = 0.837 Nm

Explanation:

Part a)

As we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(800/60)[/tex]

[tex]\omega = 83.77 rad/s[/tex]

now the angular momentum is given as

[tex]L = I\omega[/tex]

[tex]L = (0.3)(83.77)[/tex]

[tex]L = 25.13 kg m^2/s[/tex]

Part b)

angular acceleration of fan is given as

[tex]\alpha = 2\pi\frac{800/60}{30}[/tex]

[tex]\alpha = 2.79 rad/s^2[/tex]

total number of revolutions are

[tex]N = \frac{1}{4\pi}\alpha t^2[/tex]

[tex]N = \frac{1}{4\pi}(2.79)(120^2)[/tex]

[tex]N = 3200 rev[/tex]

Part c)

Torque is given as

[tex]\tau = I\alpha[/tex]

[tex]\tau = (0.30)(2.79) = 0.837 Nm[/tex]

Birds resting on high-voltage power lines are a common sight. The copper wire on which a bird stands is 2.3 cm in diameter and carries a current of 45 A. If the bird's feet are 3.9 cm apart, calculate the potential difference between its feet. (The resistivity of the wire is 1.7 10-8 Ω · m)

Answers

Explanation:

It is given that,

Diameter of the copper wire, d = 2.3 cm

Radius of copper wire, r = 1.15 cm = 0.0115 m

Distance between bird's feet, l = 3.9 cm = 0.039 m

The resistivity of the wire, [tex]\rho=1.7\times 10^{-8}\ \Omega-m[/tex]

We need to find the potential difference between bird's feet. The resistance of the wire is calculated by :

[tex]R=\rho\times \dfrac{l}{A}[/tex]

[tex]R=1.7\times 10^{-8}\ \Omega-m\times \dfrac{0.039\ m}{\pi(0.0115\ m)^2}[/tex]

R = 0.00000159 ohms

[tex]R=1.59\times 10^{-6}\ \Omega[/tex]

Let V is the potential difference between bird's feet. It can be calculated using Ohm's law as :

[tex]V=I\times R[/tex]

[tex]V=45\ A\times 1.59\times 10^{-6}\ \Omega[/tex]

V = 0.000071 volts

or

[tex]V=7.1\times 10^{-5}\ volts[/tex]

So, the potential difference between the bird's feet is [tex]7.1\times 10^{-5}\ volts[/tex]. Hence, this is the required solution.

Even with a large estimated resistance for the bird (say, R_bird = 1000 Ω), the potential difference (V_bird) would be very small (V_bird = 45 A * 1000 Ω ≈ 45000 V).  However, in reality, most of the current will flow through the much lower resistance wire, making the actual potential difference between the bird's feet much closer to zero.

Calculation (for illustration purposes only):

Even though the potential difference is negligible, we can estimate its upper bound (worst-case scenario) by assuming all the current flows through the bird's body.

Wire Resistance:

Diameter (d) = 2.3 cm = 0.023 m

Radius (r) = d/2 = 0.0115 m

Wire length between feet (l) = 3.9 cm = 0.039 m

Resistivity (ρ) = 1.7 x 10^-8 Ω · m

Wire Resistance (R_wire) = ρ * (l / π * r^2) = (1.7 x 10^-8 Ω · m) * (0.039 m / (π * (0.0115 m)^2)) ≈ 4.2 x 10^-7 Ω (very small)

Assuming All Current Through Bird:

Current (I) = 45 A

Bird's Body Resistance (R_bird) = V_bird / I (Since we don't have V_bird, this is just a placeholder)

Total Circuit Resistance (ignoring wire resistance):

R_total = R_bird + R_wire ≈ R_bird (because R_wire is negligible)

Potential Difference Across Bird (upper bound):

V_bird = I * R_total = I * R_bird (since R_wire is negligible)

By how much does a 64.3 kg mountain climber stretch her 0.910 cm diameter nylon rope when she hangs 32.2 m below a rock outcropping? (For nylon, Y = 1.35 â 10^9 Pa. Enter your answer in centimeters.)

Answers

Answer:

23 cm

Explanation:

m = 64.3 kg, diameter = 0.910 cm

radius, r = 0.455 cm = 4.55 x 10^-3 m

A = 3.14 x (4.55 x 10^-3)^2 = 6.5 x 10^-5 m^2

L = 32.2 m

Y = 1.35 x 10^9 Pa

Let the stretched length is l.

Use the formula for Young's modulus

Y = mg L / A l

l = m g L / A Y

l = (64.3 x 9.8 x 32.2) / (6.5 x 10^-5 x 1.35 x 10^9)

l = 0.23 m

l = 23 cm

A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature of the gas is raised to 235 degrees kelvin, what is the new pressure of the gas? (Assume that there's no change in volume.) A. 854.46 kPa B. 510 kPa C. 3.636 kPa D. 1.702 kPa

Answers

Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex]

Inserting the values

[tex]\frac{400}{110}=\frac{P_{2}}{235}[/tex]

P₂ = 854.46 kPa

A car is traveling at a speed of 70 mph, a long a straight road, when the driver slams on the brakes and the car m and the car a,=- 2.00 accelerates at comes to a complete stop. How much time passes between the time when the brakes were applied and the instant the car comes to a complete stop and how far did the car travel in that time? (1609 meters 1 mile, 3600 seconds 1 hour) a.) 7.28 s and 422.10 m. b.) 15.64 s and 244.70 m. c.) 12.32 s and 428.73 m. d.) 14.53 s and 210.98 m. e.) None of the above.

Answers

Answer:

Option B is the correct answer.

Explanation:

Speed of car = 70mph [tex]=70\times \frac{1609}{3600}=31.28m/s[/tex]

Deceleration of car = 2 m/s²

We have equation of motion

          v = u + at

          0 = 31.28 - 2 t

           t = 15.64 seconds.

We also have

          [tex]s=ut+\frac{1}{2}at^2=31.28\times 15.64-\frac{1}{2}\times 2\times 15.64^2=244.70m[/tex]

Option B is the correct answer.

Calculate the heat flux (in W/m2) through a sheet of a metal 11-mm thick if the temperatures at the two faces are 350 and 110 ˚C. Assume steady-state heat flow and that the thermal conductivity of this metal is 53.0 W/m-K.

Answers

q = 1156363.6W/m².

To calculate the heat flux per unit area (W/m²) of a sheet made of metal:

q = -k(ΔT/Δx)

q = -k[(T₂ - T₁)/Δx]

Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.

With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:

q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m

q = 1156363.6W/m²

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1500 N/C just outside the surface of the sphere?

Answers

Answer:

Required charge [tex]q=2.6\times 10^{9}C[/tex].

[tex]n=1.622\times 10^{10}\ electrons[/tex]

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

[tex]E =\frac{kq}{r^2}[/tex]

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = [tex]\frac{25.0}{2}=12.5cm=0.125m[/tex]

thus, according to the given data

[tex]1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}[/tex]

or

[tex]q=\frac{0.125^2\times 1500}{9\times 10^{9}}[/tex]

or

Required charge [tex]q=2.6\times 10^{9}C[/tex].

Now,

the number of electrons (n) required will be

[tex]n=\frac{required\ charge}{charge\ of\ electron}[/tex]

or

[tex]n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}[/tex]

or

[tex]n=1.622\times 10^{10}\ electrons[/tex]

Final answer:

To produce an electric field of 1500 N/C just outside the surface of the plastic sphere, approximately 337.5 nC of excess charge must be distributed uniformly within its volume.

Explanation:

To produce an electric field just outside the surface of the sphere, the excess charge on the surface of the sphere should be uniformly distributed within its volume. The electric field just outside a uniformly charged sphere is given by:

E = k * q / r2

Where E is the electric field, k is the Coulomb constant [tex](9 * 109 N m2/C^2)[/tex] q is the charge, and r is the distance from the center of the sphere to the point outside the surface. In this case, E is given as 1500 N/C and r is half the diameter of the sphere (12.5 cm).

Substituting the values into the equation, we can solve for q:

[tex]1500 N/C = (9 * 109 N m2/C2) * q / (0.125 m)^2[/tex]

Solving for q, we find that the excess charge on the surface of the sphere should be approximately [tex]3.375 x 10-7 C,[/tex] or about 337.5 nC.

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/s at an angle of 35.0° above the table, and it lands on the magazine 0.0770s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

Answers

Answer:

Thickness of magazine = 9.42 mm.

Explanation:

Considering vertical motion of spider:-

Initial velocity, u =  0.870 sin 35 = 0.5 m/s

Acceleration , a = -9.81 m/s²

Time, t = 0.077 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 0.5 x 0.077 - 0.5 x 9.81 x 0.077²

    s = 9.42 x 10⁻³ m = 9.42 mm

Thickness of magazine = 9.42 mm.

The thickness of the magazine is determined using the vertical component of the spider's initial velocity and kinematic equations for constant acceleration. By calculating and converting the vertical displacement to millimeters, we can find the thickness of the magazine the spider lands on.

Given that the initial velocity in the vertical direction (Vy) can be found using the formula Vy = V * sin(θ), where V is the initial velocity of the spider and θ is the angle of projection, we can calculate Vy = (0.870 m/s) * sin(35.0°).

The vertical displacement (y) can then be calculated using the kinematic equation for constant acceleration, y = Vy * t + (1/2) * g * t^2, where t is the time of flight, and g is the acceleration due to gravity (approximately -9.81 m/s^2). Since we want to find the displacement in the upward direction and the spider lands on the magazine after 0.0770 s, we are looking for the magnitude of y when the spider lands.

Substituting the values we have: y = (0.870 m/s) * sin(35.0°) * 0.0770 s + (0.5) * (-9.81 m/s^2) * (0.0770 s)^2. The negative sign in the acceleration term accounts for the direction of gravity, which is downwards. After calculating this expression, we convert the result from meters to millimeters (by multiplying by 1000) to obtain the thickness of the magazine.

As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear a frequency of 65 Hz. What is the speed of the car? The speed of sound in the air is 343 m/s.

Answers

Answer:

26.8 m/s

Explanation:

[tex]v[/tex]  = constant speed of the car

[tex]V[/tex]  = speed of sound = 343 m/s

[tex]f[/tex] = actual frequency of the horn

[tex]f_{app}[/tex] = frequency heard as the car approach = 76 Hz

frequency heard as the car approach is given as

[tex]f_{app}=\frac{vf}{V - v}[/tex]

[tex]76 =\frac{vf}{343 - v}[/tex]                               eq-1

[tex]f_{rec}[/tex] = frequency heard as the car recedes = 65 Hz

frequency heard as the car goes away is given as

[tex]f_{rec}=\frac{vf}{V + v}[/tex]

[tex]65 =\frac{vf}{343 + v}[/tex]                                  eq-2

dividing eq-1 by eq-2

[tex]\frac{76}{65}=\frac{343+v}{343-v}[/tex]

[tex]v[/tex] = 26.8 m/s

Final answer:

To determine the car's speed using the Doppler Effect, we calculate the difference in observed sound frequencies as the car approaches and moves away. Applying formulas for Doppler Effect calculations, the speed of the car comes out to be approximately 14.6 m/s.

Explanation:

The question revolves around the phenomenon known as the Doppler Effect, which is observed when a sound source moves relative to an observer. To calculate the speed v of the car, we use the Doppler Effect equations for sound frequencies heard when the source is moving towards and then away from the observer:

For the source approaching:

f' = f * ((v + vo) / (v - vs))

, where:

f' is the observed frequency when the source is approaching (76 Hz)

f is the original frequency emitted by the source

v is the speed of sound (343 m/s)

vo is the speed of the observer (0 m/s, since the observer is stationary)

vs is the speed of the source (the car's speed, what we are solving for)

For the source receding:

f'' = f * ((v - vo) / (v + vs))

, where:

f'' is the observed frequency when the source is receding (65 Hz)

To find the car's speed, we need to solve for vs in both equations. By eliminating f (since it's the same for both equations), we can solve for vs. Using these equations, we determine that the speed of the car is approximately 14.6 m/s.

A 100-W light bulb generates 95W of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 0.50 mm thick. What is the temperature difference between the inner and outer surfaces of the glass? Use 'deg C' as your units.

Answers

Final answer:

To calculate the temperature difference between the inner and outer surfaces of the glass bulb in a 100 W bulb that generates 95W of heat, we use a formula from the principles of heat conduction where we input parameters including the heat generated by the bulb, the thermal conductivity of the glass, the surface area of the glass and the thickness of the glass.

Explanation:

In order to calculate the temperature difference between the inner and outer surfaces of the glass bulb, we need to use the formula for heat conduction, which is given by the formula Q = (k*A*ΔT)/d, where Q is the heat generated by the bulb, k is the thermal conductivity of the glass, A is the surface area of the glass, ΔT is the temperature difference, and d is the thickness of the glass. In this case, we know that the light bulb generates 95W of heat, the radius of the bulb is 3.0 cm, the thickness of the glass is 0.50 mm. Assuming the thermal conductivity of the glass to be 0.8 W/m.K, we can substitute these values into the formula to calculate the temperature difference ΔT = Qd / (k*A). Note that here, the surface area of the glass, A = 4πr².

As you can see, the calculation requires a clear grasp of the concepts of heat conduction, thermal conductivity, and physical constants of materials. Understanding how these factors interact is key to solving problems about heat transfer in Physics.

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If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?

Answers

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

[tex]P=\dfrac{W}{t}[/tex]

Work done is equal to the change in kinetic energy of the sprinter.

[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}[/tex]

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

Final answer:

The average power generated is 900 Watts, which is approximately 1.21 horsepower.

Explanation:

The question involves calculating the average power generated by a sprinter during acceleration. Power is defined as the work done per unit time. To find the power, we need to first calculate the work done, which, in the case of the sprinter, is the kinetic energy gained as they accelerate to the final speed.

In this case, we can calculate the kinetic energy (KE) of the sprinter using the formula KE = 0.5 × mass ×[tex]velocity^2[/tex]. The mass of the sprinter is given as 54 kg and the final velocity is 10 m/s. Thus, KE = 0.5 × 54 kg × [tex](10 m/s)^2[/tex] = 2700 Joules.

Now, the power can be found by dividing the work done by the time it takes to do that work. The sprinter takes 3 seconds to achieve this final velocity. So, the average power P is P = work/time = 2700 Joules / 3 s = 900 Watts.

Converting this power to horsepower (1 horsepower = 746 Watts), we get approximately P = 900 W × 1 hp/746 W = 1.21 horsepower.

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3

Answers

Answer:

(c) 3P/5

Explanation:

The formula to calculate the power is:

[tex]P=\frac{W}{T}[/tex]

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

[tex]P=\frac{3W}{5T}=\frac{3}{5}\frac{W}{T}=\frac{3}{5}P[/tex]

A bluebird has mass of 34 grams. It flies from the ground to the top of an 8.5-m tree. What is the change in the bluebird's gravitational potential energy as it flies to the top of the tree? 2800J 290J 2.8J 280J 0.29J

Answers

Answer:

Gravitational potential energy, PE = 2.8 J

Explanation:

It is given that,

Mass of the bluebird, m = 34 g = 0.034 kg

It flies from the ground to the top of an 8.5-m tree, h = 8.5 m

We need to find the change in the bluebird's gravitational potential energy as it flies to the top of the tree. It can be calculated as :

[tex]PE=m\times g\times h[/tex]

[tex]PE=0.034\ kg\times 9.8\ m/s^2\times 8.5\ m[/tex]

PE = 2.83 J

or

PE = 2.8 J

So, the gravitational potential energy as it flies to the top of the tree is 2.8 J. Hence, this is the required solution.

Motor oil, with a viscosity of 0.25 N ∙ s/m 2, is flowing through a tube that has a radius of 5.0 mm and is 25 cm long. The drop in pressure is 300 kPa. What is the volume of oil flowing through the tube per second?

Answers

Answer:

1.18 x 10^-3 m^3/s

Explanation:

η = 0.25 N s/m^2, radius, r = 5 mm = 0.005 m, l = 25 cm = 0.25 m

P = 300 kPa = 300,000 Pa, Volume of flow = ?

By use of Poiseuillie's equation

Volume of flow = π P r^4 / (8 η l)

Volume of flow = (3.14 x 300000 x 0.005^4) / (8 x 0.25 x 0.25)

Volume of flow = 1.18 x 10^-3 m^3/s

Final answer:

Using Poiseuille's Law for viscous flow, the volume of oil flowing through the tube per second is calculated to be 1.8 liters/sec.

Explanation:

The volume of oil flowing through the tube per second can be calculated using Poiseuille's Law, which is an equation for viscous flow. This Law states that the flow rate of fluid through a pipe (Q) is directly proportional to the fourth power of the radius (r) of the tube, the pressure difference (ΔP) and inversely proportional to the viscosity (η) of the fluid, and the length (L) of the tube.

Using Poiseuille's Law, we have:

Q = (π/8) * (ΔP/r^4L) * η

By substituting the provided values:

Q = (π/8) * (300,000 Pa/(0.005 m)^4 * 0.25 m) * 0.25 N ∙ s/m^2

This yields: Q = 0.0018 m^3/s, or 1.8 liters/sec.

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A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is found to have made an angular displacement of 400 radians. (a) How fast was the wheel spinning initially? (b) What is the final angular velocity of the wheel?

Answers

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

              u = 30 rad/s

   Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

     Here u = 30 rad/s

               t = 8 s

              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

An airplane is moving horizontally with a constant velocity of (+ 115 m/s) at altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a "care package' that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground

Answers

Answer:

The time required for the package to hit the ground = 14.63 seconds.

Explanation:

Considering vertical motion of care package:-

Initial velocity, u =  0 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 1050 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    1050 = 0 x t + 0.5 x 9.81 x t²

    t = 14.63 seconds

The time required for the package to hit the ground = 14.63 seconds.

Final answer:

To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we find that it will take approximately 14.63 seconds for the care package to hit the ground.

Explanation:

To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. Since the horizontal velocity of the airplane does not change, it will not affect the time of flight. The vertical motion can be analyzed using the equation of motion for free fall, which is h = (1/2)gt^2, where h is the initial vertical displacement, g is the acceleration due to gravity, and t is the time of flight. In this case, the initial vertical displacement is 1050 meters and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values into the equation, we can solve for t and find the time required for the care package to hit the ground.

By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we have 1050 = (1/2)(9.8)t^2. Solving for t, we get t^2 = (1050) / (0.5)(9.8) = 214.29. Taking the square root of both sides, we find t ≈ 14.63 seconds. Therefore, it will take approximately 14.63 seconds for the care package to hit the ground.

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A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center. What is the relationship between the rotational speeds of the two children?

Answers

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

[tex]f = \frac{4.04}{60} =0.067 rev/s[/tex]

here we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(0.067)[/tex]

[tex]\omega = 0.42 rad/s[/tex]

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

At the instant a current of 0.40 A is flowing through a coil of wire, the energy stored in its magnetic field is 9.5 ✕ 10−3 J. What is the self-inductance of the coil (in H)?

Answers

Final answer:

The self-inductance of the coil is calculated using the energy stored in its magnetic field and the current flowing through it. By rearranging the formula W = (1/2) * L * I², the self-inductance L is found to be 0.11875 H.

Explanation:

To calculate the self-inductance of the coil, we can use the formula for the energy stored in the magnetic field of an inductor:

Energy (W) = (1/2) * L * I²

Where:

W is the energy stored in the magnetic field

L is the self-inductance of the coil

I is the current flowing through the coil

From the question, we're given that the energy (W) is 9.5 x 10⁻³ J, and the current (I) is 0.40 A. We can rearrange the formula to solve for L:

L = 2W / I²

Substitute the given values:

L = 2 * (9.5 x 10⁻³ J) / (0.40 A)²

L = 2 * (9.5 x 10⁻³ J) / (0.16 A²)

L = (19 x 10⁻³ J) / (0.16 A²)

L = 0.11875 H

Therefore, the self-inductance of the coil is 0.11875 H (henrys).

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.

Answers

Answer:

The formula is Work = Force x distance moved

so that would be 0.600 x 5 = 3 Joules which is 0.003 kilojoules

The work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories

What is Work in physics?

Work in physics is defined as the dot product of force and the displacement produced by it.

Given is an attendant who pushed 0.600 m horizontally with a force of 5.00 N.

From the definition of work done, we can write -

W = F.d

W = Fd cosФ

Then angle between force and displacement is . Therefore, cosФ is equal to 1.

W = Fd

W = 5 x 0.6

W = 3 joules = 3/4184 = 0.00072 Kilocalories

Therefore, the work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories

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The temperature of a pot of water is 90.0 °C and the mass of the water is 112 g. A block with a mass of 21.0 g and a temperature of 20.0°C is dropped into the water. The final temperature of the water and the block is 88.3 °C. The specific heat of water is exactly 1 cal/g.°C. How much heat was lost by the water? O 1.90 x10^2 cal O 35.7 cal O 7650 cal O 1.70 cal How much heat was gained by the block? O 1430 cal O 190 x10^2 cal O 35.7 cal O 68.3 cal What is the specific heat of the block? O 0.133 cal/g.°C O 5.33 cal/g.°C O273000 cal/g.°C O0.0249 cal/g.°C

Answers

Answer:

1.90 x 10² cal

1.90 x 10² cal

0.133 cal/(g °C)

Explanation:

For water :

[tex]m_{w}[/tex] = mass of water = 112 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/(g °C)

[tex]T_{wi}[/tex] = initial temperature of water = 90.0 °C

[tex]T_{wf}[/tex] = final temperature of water = 88.3 °C

[tex]Q_{w}[/tex] = Heat lost by water

Heat lost by water is given as

[tex]Q_{w}= m_{w}c_{w}(T_{wi} - T_{wf})[/tex]

[tex]Q_{w}[/tex] = (112) (1) (90.0 - 88.3)

[tex]Q_{w}[/tex] = 1.90 x 10² cal

[tex]Q_{B}[/tex] = Heat gained by the block

As per conservation of energy

Heat gained by the block = Heat lost by water

[tex]Q_{B}[/tex] = [tex]Q_{w}[/tex]

[tex]Q_{B}[/tex] = 1.90 x 10² cal

For Block :

[tex]m_{B}[/tex] = mass of block = 21.0 g

[tex]c_{B}[/tex] = specific heat of block

[tex]T_{bi}[/tex] = initial temperature of block = 20.0 °C

[tex]T_{bf}[/tex] = final temperature of block = 88.3 °C

[tex]Q_{B}[/tex] = Heat gained by Block = 1.90 x 10² cal

Heat gained by water is given as

[tex]Q_{B}[/tex] = m_{B}c_{B}(T_{bf} - T_{bi})[/tex]

1.90 x 10²  = (21.0) (88.3 - 20.0) c_{B}

c_{B} = 0.133 cal/(g °C)

The velocity field of a steady flow is given as V=(U0+bx)i?b y j , where U0 and b are constant. Find the acceleration in the x- and y-directions if U0=1m/ s and b=0.3m .

Answers

Explanation:

The velocity field o a steady flow is given by :

[tex]V=(U_o+bx)i[/tex]

Where

U₀ and b are constant

We know that, the acceleration is given by :

[tex]a=\dfrac{dV}{dt}[/tex]

[tex]a=\dfrac{d((U_o+bx)i)}{dt}[/tex]

a = b i

a = 0.3 m/s²

So, acceleration corresponding to this velocity is 0.3 m/s². Hence, this is the required solution.

An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?

Answers

Answer:

0.686 Ohm

Explanation:

Heat energy H = 50 cal = 50 x 4.2 J = 210 J, time t = 1 second, V = 12 V

Let R be the resistance.

Heat energy = V^2 x t / R

210 = 12 x 12 x 1 / R

210 = 144 / R

R = 144 / 210 = 0.686 Ohm

The resistance of the heater wire is approximately 0.688 ohms.

To calculate the resistance of the heater wire, we need to use the information that 50 cal of heat is generated per second when a potential difference of 12 V is applied across the leads. The first step is to convert the heat from calories to joules, which is the standard unit of energy in Physics. Since 1 calorie is equivalent to 4.184 joules, we can calculate the power (P) in joules per second (or watts) by multiplying the heat generated per second by this conversion factor:

P = 50 cal/s ×4.184 J/cal = 209.2 J/s = 209.2 W

Now, we can use the formula for electric power P = V^2/R, where P is the power, V is the potential difference, and R is the resistance. By rearranging the formula to solve for R, we get R = V^2/P.

Plugging our values into this equation gives us:

R = (12 V)^2 / (209.2 W) = 144 V^2 / 209.2 W = approx 0.688 ohms.

An incandescent lightbulb emits 96 W of radiation. If the filanent is at a temperature of 3242 K, what is the area of the filament

Answers

Answer:

1.53 x 10^-5 m^2

Explanation:

use the Stefan's law

Energy per unit time = σ x A x T^4

σ = 5.67 x 10 -8 W/m^2 K^4

96 = 5.67 x 10^-8 x A x (3242)^4

A = 1.53 x 10^-5 m^2

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