Which of the following is a property of an object not a result of something happening to the object? Select one: weight a. b. inertia C. momentum d. force

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

[tex]a_{t}[/tex] = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

[tex]a_{r}[/tex] = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

[tex]a = \sqrt{a_{t}^{2} + a_{r}^{2}}[/tex]

[tex]a = \sqrt{0.18^{2} + 0.38^{2}}[/tex]

[tex]a[/tex] = 0.42 m/s²

The magnitude of the resultant acceleration of the point on the flywheel's rim after it has turned through 60.0° is approximately 0.424 m/s².

To compute the magnitude of the resultant acceleration of a point on the flywheel's rim, we need to use the equation:

θ = ω0t + 0.5αt2

Where θ is the angle turned, ω0 is the initial angular velocity, and α is the angular acceleration.

Plugging in the given values, the equation becomes:

60.0° = 0 + 0.5 * 0.600 rad/s² * t2

Simplifying, we get:

t2 = (60.0° * 2) / 0.600 rad/s²

t2 = 0.200 rad/s²

Therefore, the magnitude of the resultant acceleration is:

a = ωt = 0.600 rad/s² * sqrt(0.200 rad/s²) ≈ 0.424 m/s²

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Answer:

output work is not possible to have more than 294 kJ value so this is not reasonable claim

Explanation:

As we know that the efficiency of heat engine is given as

[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]

now we will have

[tex]T_2 = 290 K[/tex]

[tex]T_1 = 500 K [tex]

[tex]\eta = 1 - \frac{290}{500}[/tex]

[tex]\eta = 0.42[/tex]

now we know that efficiency is defined as

[tex]\eta = \frac{Work}{Heat}[/tex]

[tex]0.42 = \frac{W}{700}[/tex]

[tex]W = 294 kJ[/tex]

So output work is not possible to have more than 294 kJ value

To determine if the inventor's claim of developing a heat engine is reasonable, we can calculate the maximum theoretical efficiency of the engine and compare it to the claimed net work output and heat input.

A heat engine operates between a hot reservoir and a cold reservoir, absorbing heat from the hot reservoir and converting some of it into work, while rejecting the remaining heat to the cold reservoir. The efficiency of a heat engine is determined by the temperature of the reservoirs. In this case, the inventor claims that the heat engine receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting waste heat to a sink at 290 K. To determine if this claim is reasonable, we can calculate the theoretical maximum efficiency of the heat engine using the Carnot efficiency formula.

The Carnot efficiency formula is given by:

Efficiency = 1 - (Tc / Th)

where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this case, Tc = 290 K and Th = 500 K.

Plugging in these values into the formula, we have:

Efficiency = 1 - (290 K / 500 K) = 1 - 0.58 = 0.42 or 42%

Therefore, the maximum theoretical efficiency for a heat engine operating between these temperatures is 42%. Since the claimed net work output of the heat engine is 300 kJ, we can calculate the maximum heat input by dividing the net work output by the efficiency:

Maximum heat input = Net work output / Efficiency = 300 kJ / 0.42 = 714.3 kJ

Since the claimed heat input is 700 kJ, which is slightly less than the maximum calculated heat input, it is reasonable to say that the inventor's claim is possible.

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I believe the answer is A: True

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

[tex]\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years[/tex]

∴ Time taken by New York to move 10° west is 55000000 years

Explanation:

The Coulomb's law states that the force acting on two charges is directly proportional to the product of charges and inversely proportional to the square of distance between them . Mathematically, it is given by

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Where

k is the electrostatic constant

q₁ and q₂ are charges

r is the distance between them.

The SI unit of electric force is Newton. It can be attractive or repulsive. The attraction or repulsion depend on charges. If both charges are positive, the force is repulsive and if both are opposite charges, the force is attractive.

Answer:

[tex]Q = 4.63 \times 10^6 J[/tex]

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

[tex]s = 240 J/kg C[/tex]

now we will have

[tex]Q = ms\Delta T + mL[/tex]

[tex]\Delta T = 961.8 - 20 [/tex]

[tex]\Delta T = 941.8 degree C[/tex]

now from above equation

[tex]Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)[/tex]

[tex]Q = 3.1 \times 10^6 + 1.53 \times 10^6[/tex]

[tex]Q = 4.63 \times 10^6 J[/tex]

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

Answer:

Explanation:

There are two types of collision.

(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.

In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The total mechanical energy of the system before collision = total mechanical of the system after the collision

The rate of cooling provided by a reversible refrigerator, working with a 10-kW compressor and thermal energy reservoirs at 250 K and 310 K, is calculated using the refrigerator's Carnot coefficient and the power of the compressor. With these specific conditions, the cooling rate is calculated to be 50 kW.

The question pertains to the functioning of a reversible refrigerator, specifically the cooling rate provided by the appliance which uses a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. To determine the cooling rate, we must first understand some basics about the refrigerator's operation.

A reversible refrigerator absorbs heat Qc from a cold reservoir and discards it into a hot reservoir, while work W is done on it. This work is represented by the power exerted by the compressor. The refrigerator functions by moving a coolant through coils, which absorbs heat from the contents of the refrigerator and releases it outside.

The coefficient of performance or Carnot coefficient (KR) of the refrigerator can be computed using the formula KR = Tc / (Th - Tc) where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this situation, KR = 250 / (310 - 250) = 5.

Since the work done W is the compressor power P, we can use the formula Qc = KR * P to find the cooling rate. Substituting the known values, Qc = 5 * 10 kW = 50 kW. Therefore, the rate of cooling provided by this refrigerator is 50 kW.

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Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ [tex]ft^{3}[/tex] to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = [tex]\pi r^{2}[/tex] =  [tex]\pi 4^{2}[/tex] =16[tex]\pi[/tex] [tex]ft^{2}[/tex]

Now,

Work done required = [tex]\int_{0}^{8} 52\times 16\pi (21 - y)dy[/tex]

                                  = [tex]832\pi \int_{0}^{8} (21 - y)dy[/tex]

                                  = 832[tex]\pi([/tex][tex][ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\[/tex])

                                  = 113152[tex]\pi[/tex] = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

Answer:

5.44×10⁶ m

Explanation:

For a satellite with period t and orbital radius r, the velocity is:

v = 2πr/t

So the centripetal acceleration is:

a = v² / r

a = (2πr/t)² / r

a = (2π/t)² r

This is equal to the acceleration due to gravity at that elevation:

g = MG / r²

(2π/t)² r = MG / r²

M = (2π/t)² r³ / G

At the surface of the planet, the acceleration due to gravity is:

g = MG / R²

Substituting our expression for the mass of the planet M:

g = [(2π/t)² r³ / G] G / R²

g = (2π/t)² r³ / R²

R² = (2π/t)² r³ / g

R = (2π/t) √(r³ / g)

Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:

R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)

R = 5.44×10⁶ m

Notice we didn't need to know the mass of the satellite.

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

            [tex]a=\frac{445000}{300000}=1.48m/s^2[/tex]

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

            v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

After applying the deceleration due to the braking force for 10 seconds, the speed of the Boeing 777 airplane reduces to 12.17 m/s.

Calculating the Final Speed of a Boeing 777 After Braking

To determine the speed of a Boeing 777 aircraft after a 10-second interval of braking, we will use the relationship between force, mass, initial velocity, and acceleration provided by Newton’s second law of motion. The braking force applied on the aircraft is 445,000 N acting opposite to the direction of motion. We can calculate the deceleration using the formula:

F = ma

Where F is the force, m is the mass, and a is the acceleration (deceleration in this case since the force is applied opposite to the direction of motion). The mass (m) of the airplane is 300,000 kg, so we can solve for 'a' as follows:

a = F / m = 445,000 N / 300,000 kg = 1.483 m/s² (deceleration)

Next, we use the kinematic equation to find the final velocity (vf) after 10 seconds:

vf = vi + at

Where vi is the initial velocity, and t is the time. We know that the initial velocity (vi) is 27.0 m/s and the time (t) is 10.0 s.

Plugging the numbers in:

vf = 27.0 m/s - (1.483 m/s² × 10.0 s) = 27.0 m/s - 14.83 m/s = 12.17 m/s

Therefore, the speed of the Boeing 777 airplane 10 seconds later is 12.17 m/s.

Answer:

(a) the constant of the spring is 27095.375 J/m2

(b) 0.052 m/s

(c) F=17341.04 N

Explanation:

Hello

The law of conservation of energy states that the total amount of energy in any isolated system (without interaction with any other system) remains unchanged over time.

if we assume that there is no friction, then, the kinetic energy of the train (due to movement) will be equal to the energy accumulated in the spring

Step 1

energy of the train (kinetic)

[tex]E_{k}=\frac{m*v^{2} }{2}\\  E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J\\ \\[/tex]

step 2

energy of the spring

[tex]E_{s}=\frac{Kx^{2} }{2}\\[/tex]

where K is the constant of the spring and x the length compressed.

[tex]E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} } \\\\[/tex]

(a) the constant of the spring is 27095.375 J/m2

(b)

[tex]x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2}  }{2}\\ E_{s}=5549.1328 J\\[/tex]

equal to train energy

[tex]E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s} \\\\[/tex]

(b) 0.052 m/s

(c)

[tex]F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04[/tex]

(c) F=17341.04 N

I hope it helps

The force constant of the spring is 38.03 N/m. If the train compressed the spring 0.640 m, the train would be going at a speed of 0.418 m/s. The force the spring exerts when compressed 0.640 m is 24.34 N.

To solve these questions we use conservation of energy principles and the spring equation, F = k*x. The initial kinetic energy of the train is converted into potential energy in the spring at the point of maximum compression.

(a) What is the force constant of the spring?
Here we equate kinetic energy with potential energy by using the equations KE = 0.5 * m * v^2 and PE = 0.5 * k * x^2. We can solve for k to get k = m * v^2 / x^2, so it's (2.15 x 10^5 kg * (0.710 m/s)^2)/(2.00 m)^2 = 38.03 N/m.

(b) What speed would the train be going if it only compressed the spring 0.640 m?
Here we rearrange the previous equation for velocity, v = sqrt(k * x^2 / m) which we then place values into to calculate v = sqrt(38.03 N/m * (0.640 m)^2)/(2.15 x 10^5 kg) = 0.418 m/s.

(c) What force does the spring exert when compressed 0.640 m?
Here we refer back to Hooke's Law (F = k*x), which states that the force required to compress or extend a spring by some distance x is proportional to that distance. The spring constant (k) is the proportionality constant in this relationship. So, the force is F = (38.03 N/m * 0.640 m) = 24.34 N.

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Answer:

97.96 [tex]\frac{rad}{s}[/tex]

Explanation:

The initial angular velocity is [tex]w_{0}[/tex] = 126 rad / s.

The constant angular deceleration is 5.00 rad / s2.

The constant angular deceleration is, by definition: dw / dt.

[tex]\frac{dw}{dt}=-5 \frac{rad}{s^{2} }[/tex]

Separating variables

[tex]dw=-5 dt[/tex]

Integration (limits for w: 0 to W0; limits for t: 0 to t)

[tex]w= w_{0}-5t[/tex]

W is by definition [tex]\frac{d\alpha }{dt}[/tex], where [tex]\alpha[/tex] is the angle.

[tex]\frac{d\alpha}{dt}=w_{0} -5t[/tex]

Separating variables

[tex]d\alpha=(w_{0} -5t )dt[/tex]

Integration (limits for [tex]\alpha[/tex]: 0 to 628; limits for t: 0 to t)

[tex]\alpha =w_{0}t-(\frac{5}{2})t^{2}[/tex]

[tex]128=126t-(\frac{5}{2})t^{2}[/tex]

Put in on the typical form of a quadratic equation:

[tex]\frac{5}{2}t^{2}-126t+628=0[/tex]

Solve by using the quadratic equation formula and discard the higher result because it lacks physical sense.

t=5.608 s

Evaluate at this time the angular velocity:

[tex]w(t=5.608)=126-5*5.608[/tex]

[tex]w(5.608)=97.96 \frac{rad}{s}[/tex]

To find the final angular speed of a rotating wheel given the initial speed, angular deceleration and the angle turned, we use an equation for angular motion. We substitute the given values and solve the equation to find the final angular speed.

To solve this question, we will be using the equation for angular motion,

ω² = ω0² + 2αθ

, where ω is the final angular speed we want to find, ω0 = 126 rad/s is the initial angular speed, α = -5.00 rad/s² is the angular deceleration, and θ = 628 rad is the angle turned through. This equation is analogous to the equation for linear motion, v² = u² + 2as, where v is final velocity, u initial velocity, a acceleration and s distance. We substitute the given values into the equation to find the final angular speed. Keep in mind, because we are dealing with deceleration, our α value is a negative.

Solve the equation: ω² - ω0² = 2αθ, which gives: ω² = (ω0² + 2αθ), ω = sqrt((ω0² + 2αθ)), ω = sqrt((126rad/s)² + 2*(-5.00 rad/s²)*628 rad). So, when you calculate the square root of the total, you will find the answer for the final angular speed.

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Answer:

6.93 volts

Explanation:

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

r = radius of the orbit = 2.08 x 10⁻¹⁰ m

V = Electric potential due to proton on electron

Electric potential due to proton on electron is given as

[tex]V = \frac{kq}{r}[/tex]

Inserting the values

[tex]V = \frac{(9\times 10^{9})(1.6\times 10^{-19})}{2.08\times 10^{-10}}[/tex]

V = 6.93 volts

Final answer:

The electric potential due to the proton on an electron in an orbit with a radius of 2.08 x 10^-10 m in a hydrogen atom model is approximately -13.6 volts.

Explanation:

In the context of the model of a hydrogen atom, the electric potential due to the proton on an electron in an orbit with radius 2.08 x 10-10 m can be calculated using the formula for the electric potential V due to a point charge, which is V = k * Q / r. Here, k is the Coulomb's constant (approximately 8.99 x 109 N m2/C2), Q is the charge of the proton (1.602 x 10-19 C), and r is the distance from the proton to the electron, which is the radius of the orbit.

To find the electric potential, simply plug these values into the formula:

V = (8.99 x 109 N m2/C2) * (1.602 x 10-19 C) / (2.08 x 10-10 m)

After calculating, the electric potential is found to be approximately -13.6 volts, with the negative sign indicating that the potential energy associated with the electron is negative, which is common for bound states such as electrons in an atom.

Answer:

4.89 seconds

Explanation:

The time period of a pendulum is given by

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

For a second pendulum on earth, T = 2 second

[tex]2 = 2\pi \sqrt{\frac{l}{g}}[/tex]     ...... (1)

Now the time period is T when the pendulum is taken to moon and gravity at moon is 1/6 of gravity of earth

[tex]T = 2\pi \sqrt{\frac{l}{\frac{g}{6}}}[/tex]    ...... (2)

Divide equation (2) by equation (1)

[tex]\frac{T}{2} = \sqrt{6}[/tex]

T = 4.89 seconds

Answer:

a) [tex]t=3.199 seconds[/tex]

b) [tex]h = 11.97 m[/tex]

Explanation:

Since this problem belongs to the concept of projectile motion

a) we know,

[tex]Vcos\theta=\frac{R}{t}[/tex]

Where,

V = initial speed

Θ = angle with the horizontal

R = horizontal range

t = Time taken to cover the range 'R'

Given:

V = 27m/s

R = 60m

Θ = 46°

thus,

the equation becomes

[tex]27\times cos46^o=\frac{60}{t}[/tex]

or

[tex]t=\frac{60}{27\times cos46^o}[/tex]

[tex]t=3.199 seconds[/tex]

b)The formula for height is given as:

[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]

where,

g = acceleration due to gravity = 9.8m/s²

substituting the values in the above equation we get

[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]

or

[tex]h = 62.124-50.14[/tex]

or

[tex]h = 11.97 m[/tex]

It takes 2.22 seconds for the T-shirt to reach the fan. The fan is located at a height of 24.57 meters from the ground.

To find the time it takes for the T-shirt to reach the fan, we need to solve for the time in the horizontal motion. The horizontal distance from the gun to the fan is given as 60 m. Since the horizontal motion is constant velocity, we can use the equation d = vt and solve for t. Plugging in the values, we get t = 60 m / 27 m/s = 2.22 s.

To find the height of the fan from the ground, we need to solve for the vertical motion. The equation for vertical motion is y = yt + (1/2)gt^2, where y is the vertical displacement, yt is the initial vertical velocity, and g is the acceleration due to gravity. In this case, the vertical displacement is unknown, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values and solving for y, we get y = (1/2)(9.8 m/s^2)(2.22 s)^2 = 24.57 m.

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Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

Final answer:

By applying Newton's second law and accounting for the forces acting on the clarinet case, it is found to experience a downward vertical acceleration of 0.489 m/s², due to the net force acting on it being in the downward direction.

Explanation:

To find the vertical acceleration of the clarinet case, first identify the forces acting on it. The force of gravity (weight) pulls it downward, which can be calculated by multiplying the mass of the case by the acceleration due to gravity (9.81 m/s²). The equation for weight is W = mg, where m is mass and g is gravity.

Substituting the given values, W = 3.230 kg * 9.81 m/s² = 31.68 N. The net force [tex]F_{net}[/tex] acting on the case is the upward force by the clarinetist subtracting the weight of the case, [tex]F_{net}[/tex] = 30.10 N - 31.68 N = -1.58 N. Applying Newton's second law, F = ma, and solving for acceleration (a), a = [tex]F_{Net/m}[/tex], we find the case's vertical acceleration as a = -1.58 N / 3.230 kg = -0.489 m/s².

Therefore, the case experiences a downward acceleration of 0.489 m/s², indicating that it is slowing in its ascent or accelerating downward.

Answer:

The distance the block will slide before it stops is x= 3.4 m .

Explanation:

m1= 0,02 kg

V1= 400 m/s

m2= 2 kg

V2= ?

g= 9.8 m/s²

μ= 0.24

N = m2 * g = 19.6 N

Fr= μ * N

Fr= 4.704 N

due to the conservation of the amount of movement:

m1*V1 = m2*V2

V2= 4 m/s

Fr = m2*a

a= Fr/m2

a= -2.352 m/s²

matching momentum and amount of movement

Fr*t=m2*V2

t= 1.7 sec

x= V2*t - (a*t²)/2

x= 3.4 m

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

[tex]v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f[/tex]

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

[tex]S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2[/tex]

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

Answer:

[tex]F_c = 1.4 \times 10^6 N[/tex]

Explanation:

As we know

[tex]f = 24 rpm[/tex]

so we will have

[tex]f = 24 \frac{1}{60} hz = 0.4hz[/tex]

now angular frequency is given as

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

[tex]\omega = 0.8\pi[/tex]

Now the inwards is given as centripetal force

[tex]F_c = m\omega^2 r[/tex]

[tex]F_c = (12000)(0.8\pi)^2(\frac{37}{2})[/tex]

[tex]F_c = 1.4 \times 10^6 N[/tex]

Final answer:

The inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

Explanation:

To calculate the inward force necessary to provide each blade's centripetal acceleration, we can use the formula for centripetal force:

[tex]Fc = m \times \omega^2 \times r[/tex]

where Fc is the centripetal force, m is the mass of the blade, ω is the angular velocity, and r is the radius of the blade.

In this case, the mass of the blade is 12,000 kg, the angular velocity is 24 rpm (which can be converted to radians per second by multiplying by 2π/60), and the radius of the blade is 37 m.

Plugging these values into the formula, we get:

Fc = 12000 kg * (24 * 2π/60)^2 * 37 m

Fc ≈ 4,755,076 N

So, the inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

Answer:

23.5 J/Kg °C

Explanation:

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree Celcius.

c = Q / m ΔT

Here, Q / m = 282 J/kg, ΔT = 12

So, the specific heat = 282 / 12 = 23.5 J/Kg °C

Answer:

7.55 m/s, 6.56 m/s

Explanation:

v = 10 m/s, theta = 41 degree

Horizontal component of velocity = v x Cos theta = 10 x Cos 41 = 7.55 m/s

Vertical component of velocity = v x Sin theta = 10 x Sin 41 = 6.56 m/s

Final answer:

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

Explanation:

The student's question about the initial horizontal and vertical velocities of a volleyball serve involves breaking down the initial velocity into its component parts using trigonometric functions. Given an initial velocity (v) of 10 m/s and an angle (θ) of 41 degrees from the horizontal, the horizontal component (vx) is calculated using cosine, and the vertical component (vy) is calculated using sine.

Using the formula:

vx = v ∙ cos(θ)

vy = v ∙ sin(θ)

For this problem:

vx = 10 m/s ∙ cos(41°)

vy = 10 m/s ∙ sin(41°)

Plugging in the values yields:

vx ≈ 7.57 m/s

vy ≈ 6.46 m/s

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.

v^2 = 2×g(r - r0)

v = \sqrt{2×(-9.8)×(4 - 10)}

  = 10.84 m/s

therefore, the velocity at r = 4 meters is 10.84 m/s

The speed of the electron when it reaches point B, which is 0.395 m east of point A, is approximately 3.86×10⁵ m/s, directed towards the east.

The initial speed of the electron is provided as 4.52×10⁵ m/s. Given the electric field, we can calculate the force on the electron as F = qE, where q is the charge of the electron (-1.60 × 10⁻¹⁹ C) and E is the electric field (1.55 N/C). Hence, the force acting on the electron is F = -1.60 × 10⁻¹⁹ C * 1.55 N/C = -2.48 × 10⁻¹⁹ N.

Using F = ma, we can calculate the acceleration of the electron. Knowing the mass of the electron is 9.11 × 10⁻³¹ kg, the acceleration is a = F/m = -2.48 × 10⁻¹⁹ N / 9.11 × 10⁻³¹ kg = -2.72 × 10¹¹ m/s².

Given the distance of the movement is 0.395 m, we can use the equation v² = u² + 2as to solve for the final velocity 'v', where 'u' is the initial velocity, 'a' is the acceleration and 's' is the distance. Substituting the known values, we get v = sqrt((4.52×10⁵ m/s)² - 2 * 2.72 × 10¹¹ m/s² * 0.395 m) ≈ 3.86 x 10⁵ m/s (approximately).

So, the speed of the electron when it reaches point B is approximately 3.86 x 10⁵ m/s, directed towards the east.

https://brainly.com/question/32257278

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Answer:

Its potential energy decreases and itselectricpotential increases.

Explanation:

The electric potential does not depend on the charge but only on the magnitude of the electric field. In particular:

- Electric potential decreases when moving in the same direction of the electric field lines

- Electric potential increases when moving in the opposite direction to the field lines

So in this case, since the electron is moving in a direction opposite to the field, the electric potential increases.

However, the electric potential energy of a charge is given by

[tex]U=qV[/tex]

where

q is the charge

V is the electric potential

Here we said that the electric potential is increasing: however, the charge q of an electron is negative. This means that the product (qV) is increasing in magnitude but it is negative, so the potential energy of the electron is decreasing.

For an electron moving in a direction opposite to the electric field its potential energy decreases and its electric potential increases.

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field.it is denoted by U.

However, the electric potential energy of a charge is given by

[tex]\rm{U=qv}[/tex]

q is denoted for the charge.

V is denoted for the electric potential.

The above relation shows that the electric potential is inversely proportional to the electric potential energy.

The electric potential is independent of the charge. but it depends on the direction in which the charge is moving. If the charge moves in the direction of the electric field electric potential decreases and vice versa.

Here the electron is moving in a direction opposite to the field, the electric potential increases while electric potential energy decreases.

To learn more about the electric potential energy refers to the link.

https://brainly.com/question/12707371

The magnetic force on the wire has an x-component of 0 N, a y-component of 0.004056224 N in the negative y-direction, and a z-component of 0.000143712 N in the positive z-direction.

The question involves calculating the magnetic force on a current-carrying wire in a magnetic field using the right-hand rule and the formula F = I × (L × B), where F is the force, I is the current, L is the length of the wire, and B is the magnetic field. We know the current I is 0.720 A, the length L is 73.6 cm (which we will convert to meters), and the magnetic field components are By = 0.000270 T and Bz = 0.00770 T.

First, let's convert the length of the wire from centimeters to meters: L = 73.6 cm = 0.736 m.

The force on the wire in the x, y, and z directions (Fx, Fy, Fz) can be calculated using the cross product of the current direction (along the x-axis) and the magnetic field components. Since there is no x component for the magnetic field (Bx = 0), the force in the x-direction (Fx) will be zero.

Using the right-hand rule, the force in the y-direction (Fy) will be:

Fy = I × (L × Bz) = 0.720 A × (0.736 m × 0.00770 T) = 0.004056224 N, pointing in the negative y-direction (since the current is in the positive x-direction and Bz is positive).

Similarly, the force in the z-direction (Fz) is calculated as:

Fz = I × (L × By) = 0.720 A × (0.736 m × 0.000270 T) = 0.000143712 N, pointing in the positive z-direction.

Answer:

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 =  1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

Answer:

0.51

Explanation:

m = mass of the book = 3.5 kg

F = force applied by the broom on the book = 21 N

a = acceleration of the book

v₀ = initial speed of the book = 0 m/s

v = final speed of the book = 1.2 m/s

d = distance traveled = 0.74 m

Using the equation

v² = v₀² + 2 a d

1.2² = 0² + 2 a (0.74)

a = 0.973 m/s²

f = kinetic frictional force

Force equation for the motion of the book is given as

F - f = ma

21 - f = (3.5) (0.973)

f = 17.6 N

μ = Coefficient of kinetic friction

Kinetic frictional force is given as

f = μ mg

17.6 = μ (3.5 x 9.8)

μ = 0.51

Answer:

[tex]v' = 2.83 m/s[/tex]

Explanation:

Velocity of wave in stretched string is given by the formula

[tex]v = \sqrt{\frac{T}{\mu}}[/tex]

here we know that

T = 4 N

also we know that linear mass density is given as

[tex]\mu = 1 kg/m[/tex]

so we have

[tex]v = \sqrt{\frac{4}{1}} = 2 m/s[/tex]

now the tension in the string is double

so the velocity is given as

[tex]v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s[/tex]

[tex]v' = 2.83 m/s[/tex]

Answer:

655128 ohm

Explanation:

C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F  

V₀ = Voltage of the battery = 9 Volts  

V = Potential difference across the battery after time "t" = 4.20 Volts  

t = time interval = 3.21 sec  

T = Time constant

R = resistance  

Potential difference across the battery after time "t" is given as  

[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]

[tex]4.20 = 9 (1-e^{\frac{-3.21}{T}})[/tex]

T = 5.11 sec  

Time constant is given as  

T = RC  

5.11 = (7.8 x 10⁻⁶) R  

R = 655128 ohm