A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.20 m/s. The coefficient of kinetic friction between sled and ice is 0.115. Use energy considerations to find the distance the sled moves before it stops. m

Answer:

[tex]INTNESITY = 0.88W/m^{2}[/tex]

Explanation:

given data:

power P =  40W

Distance of light bulb from screen is  R  = 1.9 m

The light intensity considered as  the light energy falling on the surface per unit area per unit time

Intensity of light can be determined by using following formula

[tex]intesnity = \frac{P} {4\pi R^{2}}[/tex]

[tex]intesnity = \frac{40} {4\pi *1.9^{2}}[/tex]

[tex]INTNESITY = 0.88W/m^{2}[/tex]

Answer:

[tex]v = 1.11 \times 10^3 m/s[/tex]

Explanation:

By energy conservation law we will have

[tex]U_i + KE_i = U_f + KE_f[/tex]

as we know that as the projectile is rising up then due to gravitational attraction of earth it will slow down

At the highest position the speed of the projectile will become zero

So here we will have

[tex]-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0[/tex]

[tex]-\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(100)}{6.37 \times 10^6} + \frac{1}{2}(100)v^2 = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{-11})(100)}{(6.37 \times 10^{6} + 65000)}[/tex]

[tex] - 6.25 \times 10^7 + 0.5 v^2 = -6.19 \times 10^7[/tex]

[tex]v = 1.11 \times 10^3 m/s[/tex]

Answer:

Part a)

[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]

Part b)

[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]

Explanation:

Part a)

As we know that initially the two blocks are connected by a spring and initially stretched by some amount

Since the two blocks are at rest initially so its initial momentum is zero

since there is no external force on this system so final momentum is also zero

[tex]m_Av_{1i} + m_Bv_{2i} = m_Av_A + m_Bv_B[/tex]

now for initial position the speed is zero

[tex]0 = m_Av_A + m_Bv_B[/tex]

now we have

[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]

Part b)

now for ratio of kinetic energy we know that the relation between kinetic energy and momentum is given as

[tex]K = \frac{P^2}{2m}[/tex]

now for the ratio of energy we have

[tex]\frac{K_A}{K_B} = \frac{P^2/2m_A}{P^2/2m_B}[/tex]

since we know that momentum of two blocks are equal in magnitude so we have

now we have

[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]

Final answer:

To calculate the ratios of velocities and kinetic energies for two blocks connected by a massless spring released from rest, use conservation of momentum and the kinetic energy formula. The velocity ratio is mB/mA and the kinetic energy ratio is (mB/mA)².

Explanation:

The question is about two blocks connected by a massless spring on a frictionless surface. When they are released from rest after stretching the spring, we want to find the ratio of their velocities and kinetic energies during their motion.

Ratio of Velocities

The ratio of velocities is obtained through conservation of momentum. Since no external forces are involved and the spring does not have mass, the momentum of the system is conserved. Assuming block A and block B move in opposite directions after being released, their velocities will be in different directions but they will have the same magnitude of momentum.

Momentum conservation dictates that mAvA = mBvB, where vA and vB are their speeds respectively. Thus, the ratio of their velocities vA/vB = mB/mA.

Ratio of Kinetic Energies

The kinetic energy of each block is given by KE = (1/2)[tex]mv^2[/tex]. The ratio of their kinetic energies KEA/KEB is therefore [tex](m_Av_A^2)/(m_Bv_B2^)[/tex]. Inserting the ratios found in the velocities section, we find that the ratio of the kinetic energies is [tex](mB/mA)^2[/tex].

Using the kinematic equations of motion, we can calculate that a projectile launched at an initial speed of 1360 m/s from the moon's surface will reach two-fifths its initial speed at an altitude of roughly 680 kilometers.

A projectile launched vertically from the moon's surface decelerates due to the moon's gravity at a rate of 1.6 m/s². Given the initial speed of the projectile is 1360 m/s, we want to find out the altitude at which the speed is two-fifths of this initial velocity. To find out, we use the kinematic equation v² = u² + 2as.

Substituting the values into the equation, we get (2/5 * 1360)² = (1360)² - 2 * 1.6 * s. Solving this equation gives us the distance s, where s comes out to be approximately 680000 meters or 680 km.

This means that the projectile's speed is two-fifths its initial value at an altitude of approximately 680 km from the moon's surface.

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To find the altitude at which the projectile's speed is two-fifths its initial value, the loss of kinetic energy must equal the gain in potential energy. Solving for the altitude (h), we find the answer.

This problem involves physics concepts of projectile motion and gravitational acceleration. Given that the projectile is launched with an initial velocity of 1360 m/s, and we want to determine the altitude at which it has slowed to two-fifths of that speed, we will be considering the effects of the Moon's gravity on the deceleration of the projectile.

Using the formula for kinetic energy K.E = 1/2 m v^2, when the projectile's velocity slows to two-fifths of its original speed, the kinetic energy will be four-fifths less since kinetic energy is proportional to the square of the speed. This loss of kinetic energy must equal the gain in potential energy (since energy is conserved), which is given by the formula P.E = mgh (where m is mass, g is gravitational acceleration, and h is height or altitude).

Solving P.E = K.E for h, we get the altitude at which the projectile's speed is two-fifths its initial value.

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Answer:

3.91 N/C

Explanation:

u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s

Let a be the acceleration.

Use second equation of motion

s = u t + 1/2 a t^2

5.5 = 0 + 1/2 a (4 x 10^-6)^2

a = 6.875 x 10^11 m/s^2

F = m a

The electrostatic force, Fe = q E

Where E be the strength of electric field.

So, q E = m a

E = m a / q

E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)

E = 3.91 N/C

Answer:

The ball must be tossed to a height of 4 times the initial height H

Explanation:

We have equation of motion S = ut + 0.5at²

A juggler tosses balls vertically to height H.

That is

      H = 0 x t + 0.5 x a x t²

      H = 0.5at²

To what height must they be tossed if they are to spend twice as much time in the air.

      H' = 0 x 2t + 0.5 x a x (2t)²

      H' = 2at² = 4 H

So the ball must be tossed to a height of 4 times the initial height H

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]

[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]

[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]

x = - 1.4 m

Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

Given information-

The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].

The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].

The distance between the point 1 and point 2 is 2 meters away from the x axis.

Let the position of the point 1 is at x.

Thus the electric force on point Q1 is,

[tex]F_1=\dfrac{kQ1}{x^2}[/tex]

As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2

[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]

As the force of two is equal and opposite thus,

[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]

Thus the position of point 2 is,

[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]

Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.

Learn more about electric field here;

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Answer:

 D . 1.47 x 10⁻² Ω-m

Explanation:

L = length of the cylindrical resistor = 2 m

d = diameter = 0.1 m

A = Area of cross-section of the resistor = (0.25) [tex]\pi[/tex] d² = (0.25) (3.14) (0.1)² = 0.785 x 10⁻² m²

V = battery Voltage = 12 volts

[tex]i [/tex] = current flowing through the resistor = 3.2 A

R = resistance of the resistor

Resistance of the resistor is given as

[tex]R = \frac{V}{i}[/tex]

[tex]R = \frac{12}{3.2}[/tex]

R = 3.75 Ω

[tex]\rho[/tex] = resistivity

Resistance is also given as

[tex]R = \frac{ \rho L}{A}[/tex]

[tex]3.75 = \frac{ \rho (2)}{(0.785\times 10^{-2})}[/tex]

[tex]\rho[/tex] =  1.47 x 10⁻² Ω-m

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

[tex]F = qvB[/tex]

here we have

[tex]B = \frac{F}{qv}[/tex]

here we know that

[tex]F = 4.8 \times 10^{-13} N[/tex]

[tex]q = 1.6 \times 10^{-19} C[/tex]

[tex]v = 4 \times 10^6 m/s[/tex]

now from above equation we have

[tex]B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}[/tex]

[tex]B = 0.75 T[/tex]

The rotational kinetic energy of the pitcher's arm when throwing a softball at a speed of 139 km/h is calculated using the equation for rotational kinetic energy, and taking moment of inertia and angular velocity into account. The angular velocity is inferred from the linear speed of the ball and the distance it leaves the pitcher's hand from the pivot at the shoulder. The rotational kinetic energy is found to be approximately 1491 Joules.

This question pertains to the rotational kinetic energy of the pitcher's arm during the act of throwing a softball. By definition, the kinetic energy associated with rotational motion (rotational kinetic energy) can be given by the equation: K_rot = 0.5 * I * ω² where 'I' is the moment of inertia and 'ω' is the angular velocity.

To calculate the angular velocity 'ω', we can infer it from the linear speed of the ball when it leaves the pitcher's hand, as 'ω = v/r', v = 139 km/h = 38.6 m/s, and r = 0.600 m (distance ball leaves hand from pivot at shoulder). Therefore, 'ω' is approximately 64.4 rad/s.

Substituting 'I' and 'ω' into the equation above, we get K_rot = 0.5 * 0.720 kg*m² * (64.4 rad/s)² = 1491 Joules, which is the rotational kinetic energy of the pitcher's arm.

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Answer:

Part a)

[tex]f = 1.46 \times 10^7 Hz[/tex]

Part b)

[tex]KE = 3.87 \times 10^{-12} J[/tex]

Explanation:

Part a)

As we know that radius of circular path of a charge moving in constant magnetic field is given as

[tex]R = \frac{mv}{qB}[/tex]

now we have

[tex]v = \frac{qBR}{m}[/tex]

now the frequency of oscillator is given as

[tex]f = \frac{v}{2\pi R}[/tex]

[tex]f = \frac{qB}{2\pi m}[/tex]

[tex]f = \frac{(1.6 \times 10^{-19})(0.960)}{2\pi(1.67\times 10^{-27})}[/tex]

[tex]f = 1.46 \times 10^7 Hz[/tex]

PART b)

now for kinetic energy of proton we will have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}m(\frac{qBR}{m})^2[/tex]

[tex]KE = \frac{q^2B^2R^2}{2m}[/tex]

[tex]KE = \frac{(1.6 \times 10^{-19})^2(0.960)^2(0.740)^2}{2(1.67\times 10^{-27})}[/tex]

[tex]KE = 3.87 \times 10^{-12} J[/tex]

Answer:

[tex]\rho = 2940 kg/m^3[/tex]

Explanation:

When pendulum is in air then the tension in the string is given as

[tex]T = mg[/tex]

[tex]T = 38.7 N[/tex]

now when it is submerged in water then tension is decreased due to buoyancy force of water

so we will have

[tex]T_{water} = mg - F_b[/tex]

[tex]32 = 38.7 - F_b[/tex]

[tex]F_b = 6.7 N[/tex]

When rock is immersed in unknown liquid then tension is given by

[tex]T = mg - F_b'[/tex]

[tex]19 = 38.7 - F_b'[/tex]

[tex]F_b' = 19.7 N[/tex]

from buoyancy force of water we can say

[tex]F_b = 1000(V)(9.8) = 6.7[/tex]

[tex]V = 6.84 \times 10^{-4} m^3[/tex]

now for other liquid we will have

[tex]19.7 = \rho(6.84 \times 10^{-4})(9.8)[/tex]

[tex]\rho = 2940 kg/m^3[/tex]

To find the density of the unknown liquid, we can use Archimedes' principle and the difference in tension when the rock is immersed in water and in the unknown liquid. The formula for buoyant force can be used to calculate the density of the unknown liquid.

To find the density of the unknown liquid, we can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

In this case, the difference in tension when the rock is in air and when it is totally immersed in water represents the buoyant force acting on the rock. We can use the formula:

Buoyant force = weight of the fluid displaced

We know that the tension in the string when the rock is in water is 32.0 N and when it is in the unknown liquid is 19.0 N. Subtracting the tension in water from the tension in the unknown liquid, we get:

Tension in unknown liquid - Tension in water = Buoyant force

19.0 N - 32.0 N = -13.0 N

The negative sign indicates that the buoyant force is acting upward on the rock. Since the buoyant force is equal to the weight of the fluid displaced, we can equate the buoyant force to the weight of the unknown liquid displaced by the rock:

-13.0 N = -(density of unknown liquid) x (gravitational acceleration) x (volume of rock)

To find the density of the unknown liquid, we can rearrange the equation as follows:

Density of unknown liquid = (-13.0 N) / (9.8 m/s^2) x (volume of rock)

Plugging in the values, the density of the unknown liquid can be calculated.

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Answer:

For no friction condition there is no range of speed only on possible speed is 41.1 km/h

while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s

Explanation:

When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only

So we will have

[tex]Ncos\theta = mg[/tex]

[tex]Nsin\theta = \frac{mv^2}{R}[/tex]

so we have

[tex]tan\theta = \frac{v^2}{Rg}[/tex]

[tex]\theta = tan^{-1}\frac{v^2}{Rg}[/tex]

v = 41.1 km/h = 11.42 m/s

[tex]\theta = tan^{-1}\frac{11.42^2}{28(9.8)}[/tex]

[tex]\theta = 25.42^o[/tex]

so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn

Now in next case if the coefficient of static friction is 0.300

then in this case we have

[tex]v_{max} = \sqrt{(\frac{\mu + tan\theta}{1 - \mu tan\theta})Rg}[/tex]

[tex]v_{max} = \sqrt{(\frac{0.3 + 0.475}{1 - (0.3)(0.475)})(28 \times 9.8)}[/tex]

[tex]v_{max} = 15.75 m/s[/tex]

Similarly for minimum speed we have

[tex]v_{min} = \sqrt{(\frac{\mu - tan\theta}{1 + \mu tan\theta})Rg}[/tex]

[tex]v_{min} = \sqrt{(\frac{-0.3 + 0.475}{1 + (0.3)(0.475)})(28 \times 9.8)}[/tex]

[tex]v_{min} = 6.5 m/s[/tex]

So the range of the speed is from 6.5 m/s to 15.75 m/s

Answer:

2.5 min

Explanation:

Hello

by definition  the power   is the amount of work done per unit of time. in this case, we know  the total power and the work developed

[tex]1 Watt= \frac{Joule }{sec}[/tex]

Let

[tex]P=1200 W\\E=1.8 *10^{5}\\X=\frac{1.8 *10^{5} j }{1200 W}\\X= unknown\ time\ (sec) \\\\we\ need\ the\ answer\ in\ minutes,\\ 1\ min=60\ sec\\\\x=150 sec \\150\ sec*(\frac{1 min}{ 60 sec})=2.5 min\\[/tex]

hence , the data is not altered, we did it like this to eliminate the sec units.

Answer 2.5 min

I hope it helps

The food was in the 1200 W microwave for 150 seconds, which is equivalent to 2.5 minutes.

To calculate how long the food was in the microwave, we use the formula for power, which is the rate at which work is done or energy is transferred:

Power (P) = Energy (E) / Time (t)

First, rearrange this formula to solve for time (t):

t = E / P

Plugging in the given values:

So:

t = (1.8 times 10⁵J) / (1200 W)

Calculate this to find the time in seconds, and then convert it to minutes as follows:

t = 150 seconds

Divide by 60 to get minutes:

t = 2.5 minutes

Answer:

517.6 N

8.63 m/s²

Explanation:

M = mass of earth = 5.98 x 10²⁴ kg

m = mass of the person = 60 kg

W = weight of the person

R = radius of earth = 6.4 x 10⁶ m

d = distance of ISS above the surface of earth = 400 km = 4 x 10⁵ m

Weight of person on the ISS is given as

[tex]W = \frac{GMm}{(R+d)^{2}}[/tex]

[tex]W = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})(60)}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]

W = 517.6 N

Acceleration due to gravity is given as

[tex]g = \frac{GM}{(R+d)^{2}}[/tex]

[tex]g = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]

g = 8.63 m/s²

Answer:

The time period of the wave is 0.002 sec.

Explanation:

Given that,

Frequency = 0.5 kHz

Phase difference [tex]\phi=\dfrac{\pi}{3}[/tex]

Path difference = 1.5 cm

We need to calculate the time period

Using formula of time period

The frequency is the reciprocal of time period.

[tex]f =\dfrac{1}{T}[/tex]

[tex]T=\dfrac{1}{f}[/tex]

Where, f = frequency

Put the value into the formula

[tex]T=\dfrac{1}{0.5\times10^{3}}[/tex]

[tex]T=0.002\ sec[/tex]

Hence, The time period of the wave is 0.002 sec.

Answer:

[tex]x_{1}=-2,x_{2}=4[/tex]

value of g(x) at these points are as follows

g(-2)=30

g(4)=-78

Explanation:

Given

g(x)=[tex]x^{3}-3x^{2}-24x+2[/tex]

Differentiating with respect to x we get

[tex]g'(x)=3x^{2}-6x-24[/tex]

to obtain point of extrema we equate g'(x) to zero

[tex]g'(x)=3x^{2}-6x-24\\\\\therefore 3x^{2}-6x-24=0\\\\\Rightarrow x^{2}-2x-8=0\\\\x^{2}-4x+2x-8=0\\x(x-4)+2(x-4)=0\\(x+2)(x-4)=0[/tex]

Thus the critical points are obtained as [tex]x_{1}=-2,x_{2}=4[/tex]

The values at these points are as

[tex]g(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+2=30\\\\g(4)=(4)^{3}-3(4)^{2}-24(4)+2=-78[/tex]

Answer:

The body will strike at angle 60.46°

Explanation:

Vertical motion of body:

 Initial speed, u = 40sin30 = 20m/s

 Acceleration, a = 9.81 m/s²

 Displacement, s = 170 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 20² + 2 x 9.81 x 170

    v = 61.12 m/s

 Final vertical speed = 61.12 m/s

 Final horizontal speed = initial horizontal speed = 40cos30= 34.64m/s

 Final velocity = 34.64 i - 61.12 j m/s

 Magnitude

     [tex]v=\sqrt{34.64^2+(-61.12)^2}=70.25m/s[/tex]

 Direction

      [tex]\theta =tan^{-1}\left ( \frac{-61.12}{34.64}\right )=-60.46^0[/tex]

 The body will strike at angle 60.46°

Answer:

True

Explanation:

The force on the electron when it enters in a magnetic field is given by

F = q ( v x B)

F = -e x V x B x Sin∅

here, F is the force vector, B be the magnetic field vector and v be the velocity vector.

If the angle between the velocity vector and the magnetic field vector is 0 degree, then force is zero.

When the electrons enters in the magnetic field at any arbitrary angle, it experiences a force and hence it accelerate up.

Answer: [tex]22.8^0C[/tex]

Explanation:-

[tex]Q_{absorbed}=Q_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]      

where,

[tex]m_1[/tex] = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]

[tex]m_2[/tex] = mass of water = 21.9 kg = 21900 g

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_1[/tex] = temperature of iron horseshoe = [tex]600^oC[/tex]

[tex]T_2[/tex] = temperature of water = [tex]21.8^oC[/tex]

[tex]c_1[/tex] = specific heat of iron horseshoe = [tex]0.450J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of water =  [tex]4.184J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]

[tex]350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)][/tex]

[tex]T_{final}=22.8^0C[/tex]

Therefore, the final equilibrium temperature is [tex]22.8^0C[/tex].

Answer:

The inductance is 0.021 H.

Explanation:

Given that,

Number of turns = 16

Current = 3 A

Total flux [tex]\phi=4\times10^{-3}\ Tm^2[/tex]

We need to calculate the inductance

Using formula of total flux

[tex]N\phi = Li[/tex]

[tex]L=\dfrac{N\phi}{i}[/tex]

Where, i = current

N = number of turns

[tex]\phi[/tex] = flux

Put the value into the formula

[tex]L=\dfrac{16\times4\times10^{-3}}{3}[/tex]

[tex]L=0.021\ H[/tex]

Hence, The inductance is 0.021 H.

Answer:

(a) 0 rad

(b) 4 rad/s

(c) 28 rad/s

(d)  12 rad/s^2

(e) 6 rad/s^2

(f) 18 rad/s^2  

Explanation:

[tex]\theta (t)=4t-3t^{2}+t^{3}[/tex]   .... (1)

(a) here, we need to find angular displacement when t = 0 s

Put t = 0 in equation (1), we get

[tex]\theta (t=0)=0[/tex]

(b) Angular velocity is defined as the rate of change of angular displacement.

ω = dθ / dt

So, differentiate equation (1) with respect to t.

[tex]\omega =\frac{d\theta }{dt}=4-6t+3t^{2}[/tex]   .... (2)

Angular velocity at t = 2 s

Put t = 2 s in equation (2), we get

ω = 4 - 6 x 2 + 3 x 4 = 4 rad/s

(c) Angular velocity at t = 4 s

Put t = 4 s in equation (2), we get

ω = 4 - 6 x 4 + 3 x 16 = 4 - 24 + 48 = 28 rad/s

(d) Average angular acceleration,

[tex]\alpha =\frac{\omega (t=4s)-\omega (t=2s)}{4-2}[/tex]

α = (28 - 4) / 2 = 12 rad/s^2

(e) The rate of change of angular velocity is called angular acceleration.

α = dω / dt

α = - 6 + 6 t

At t = 2 s

α = - 6 + 12 = 6 rad/s^2

(f) At t = 4 s

α = - 6 + 24 = 18 rad/s^2

Answer:

Coefficient of static friction between the tires and the road is 0.92.

Explanation:

It is given that,

Radius of the curve, r = 120 m

Speed, v = 119 km/h = 33.05 m/s

We need to find the coefficient of static friction between the tires and the road. In a curved road the safe velocity is given by :

[tex]v=\sqrt{\mu rg}[/tex]

[tex]\mu[/tex] is the coefficient of static friction

g is acceleration due to gravity

[tex]\mu=\dfrac{v^2}{rg}[/tex]

[tex]\mu=\dfrac{(33.05\ m/s)^2}{120\ m\times 9.8\ m/s^2}[/tex]

[tex]\mu=0.92[/tex]

So, the coefficient of static friction between the tires and the road is 0.92. Hence, this is the required solution.

The angular speed of the truck's wheels is approximately 3422.35 rad/min.

The angular speed [tex](\( \omega \))[/tex] is related to the linear speed [tex](\( v \))[/tex] and the radius [tex](\( r \))[/tex] of the wheels by the formula [tex]\( v = r \cdot \omega \)[/tex].

Firstly, we need to find the radius of the wheels, which is half of the diameter. So, [tex]\( r = \frac{34}{2} = 17 \) inches[/tex].

Now, convert the speed from miles per hour to inches per minute. There are 5280 feet in a mile, 12 inches in a foot, and 60 minutes in an hour.

[tex]\[ v = 55 \, \text{mi/h} \times \frac{5280 \, \text{ft}}{1 \, \text{mi}} \times \frac{12 \, \text{in}}{1 \, \text{ft}} \times \frac{1 \, \text{h}}{60 \, \text{min}} \][/tex]

[tex]\[ v \approx 58080 \, \text{in/min} \][/tex]

Now, use the formula [tex]\( v = r \cdot \omega \)[/tex] to find [tex]\( \omega \)[/tex]:

[tex]\[ \omega = \frac{v}{r} \][/tex]

[tex]\[ \omega = \frac{58080 \, \text{in/min}}{17 \, \text{in}} \][/tex]

[tex]\[ \omega \approx 3422.35 \, \text{rad/min} \][/tex]

Therefore, the angular speed of the truck's wheels is approximately 3422.35 rad/min.

Answer:

1.59 m/s^2, 65.2°

Explanation:

F1 = 390 N North

F2 = 180 N east

m = 270 kg

Net force is the vector sum of both the forces.

[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]

[tex]F = \sqrt{390^{2}+180^{2}}[/tex]

F = 429.53 N

Direction of force

tan∅ = F1 / F2 = 390 / 180 = 2.1667

∅ = 65.2°

The direction of acceleration is same as the direction of net force.

The magnitude of acceleration is

a = F / m = 429.53 / 270 = 1.59 m/s^2

Final answer:

The magnitude of the sailboat's acceleration is 1.59 m/s², and the direction is 25 degrees east of north. This is calculated using Newton's second law and vector addition of the orthogonal forces exerted by the wind and water.

Explanation:

To calculate the magnitude and direction of the sailboat's acceleration, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Here, we combine the forces exerted by the wind and the water to obtain the net force. The net force can be calculated using vector addition, where the force due to the wind (390 N north) and the force due to the water (180 N east) are treated as orthogonal vectors.

The net force (Fnet) is the vector sum of the two individual forces. We calculate the net force using the Pythagorean theorem:

Fnet = √Fwind² + Fwater² = √390² + 180² = √152100 + 32400 = √184500 N

So, the magnitude of the net force is approximately 429.53 N. To find the acceleration (a), we use the formula:

a = Fnet / m

Substituting the mass of the sailboat (270 kg) and the net force, we get:

a = 429.53 N / 270 kg = 1.59 m/s²

To determine the direction of the acceleration, we take the arctangent of the ratio of the forces. Since arctan(180/390) equals approximately 25 degrees, this is the angle east of north.

The sailboat's acceleration has a magnitude of 1.59 m/s² and a direction of 25 degrees east of north.

Answer:

1.718 N , attractive

Explanation:

r = 0.66 m, n = 5.7 x 10^13

q1 = 5.7 x 10^13 x 1.6 x 10^-19 = 9.12 x 10^-6 C

q2 = - 5.7 x 10^13 x 1.6 x 10^-19 = - 9.12 x 10^-6 C

F = K q1 q2 / r^2

F = 9 x 10^9 x  9.12 x 10^-6 x 9.12 x 10^-6 / (0.66)^2

F = 1.718 N

As both the charges are opposite in nature, so the force between them is attractive.

Answer:

[tex]K = 0.076 W/m K[/tex]

Explanation:

Heat required to melt the complete ice is given as

[tex]Q = mL[/tex]

here we have

m = 5.0 kg

[tex]L = 3.35 \times 10^5 J/kg[/tex]

now we have

[tex]Q = (5 kg)(3.35 \times 10^5)[/tex]

[tex]Q = 1.67 \times 10^6 J[/tex]

now the power required to melt ice in 8.4 hours is

[tex]P = \frac{Q}{t} = \frac{1.67\times 10^6}{8.4 \times 3600 s}[/tex]

[tex]P = 55.4 Watt[/tex]

now by formula of conduction we know

[tex]P = \frac{KA(\Delta T)}{x}[/tex]

now we have

[tex]55.4 = \frac{K(0.73 m^2)(25 - 5)}{0.02}[/tex]

[tex]K = 0.076 W/m K[/tex]

The change in momentum of the ball is 24 kg·m/s.

The change in momentum of the ball can be found by subtracting the initial momentum from the final momentum. The initial momentum is given by the formula p1 = m * v1, where m is the mass of the ball and v1 is its initial velocity. In this case, the initial velocity is 12 m/s downward, so the initial momentum is -12 kg·m/s. The final momentum is given by the formula p2 = m * v2, where v2 is the velocity of the ball after the bounce. Since the motion after the bounce is the mirror image of the motion before the bounce, the final velocity is 12 m/s upward. Thus, the final momentum is 12 kg·m/s. Subtracting the initial momentum from the final momentum, we get the change in momentum to be 24 kg·m/s.

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The change in momentum of a 1.0-kg ball that bounces with a velocity of 12 m/s upward after striking the ground with a velocity of 12 m/s downward is 24 kg·m/s. This is calculated using the formula for momentum, p = mv, and finding the difference between initial and final momentum.

Momentum (old{p}old{) is calculated by the formula:

p = mv

Step-by-Step Explanation:

The initial momentum just before striking the ground (p₁) is:

p₁ = mass × velocity

p₁ = 1.0 kg × (-12 m/s)

p₁ = -12 kg·m/s

The final momentum just after bouncing up (p₂) is:

p₂ = mass × velocity

p₂ = 1.0 kg × 12 m/s

p₂ = 12 kg·m/s

The change in momentum (Δp) is:

Δp = p₂ - p₁

Δp = 12 kg·m/s - (-12 kg·m/s)

Δp = 12 kg·m/s + 12 kg·m/s

Δp = 24 kg·m/s

So, the change in momentum of the ball is 24 kg·m/s.

The center of mass is given by:

∑mx/∑m

m is the mass of each object

x is the position of each object

We will assign x = 0m to the 18kg mass, therefore x = 6m for the 6kg mass.

∑mx/∑m = (18×0+6×6)/(18+6) = 1.5m

The center of mass is located 1.5m away from the 18kg mass.

The total moment of inertia of the system about the center of mass is given by:

I = ∑mr²

I is the moment of inertia

m is the mass of each object

r is the distance of each object from the center of mass

We know r = 1.5m for the 18kg mass and the rod is 6m long, therefore the 6kg mass must be r = 4.5m from the center of mass.

I = 18(1.5)² + 6(4.5)²

I = 162kg×m²

Answer:

The internal energy is 73.20 J.

Explanation:

Given that,

Weight of helium = 86 mg

Temperature = 0°C

We need to calculate the internal energy

Using formula of internal energy

[tex]U =nc_{v}T[/tex]

Where, [tex]c_{v}[/tex] = specific heat at constant volume

He is mono atomic.

So, The value of [tex]c_{v}= \dfrac{3}{2}R[/tex]

now, 1 mole of Helium = 4 g helium

n =number of mole of the 86 mg of helium

[tex]n = \dfrac{86\times10^{-3}}{4}[/tex]

[tex]n =2.15\times10^{-2}\ mole[/tex]

T = 0°C=273 K

Put the value into the formula

[tex]U = 2.15\times10^{-2}\times\dfrac{3}{2}\times8.314\times273[/tex]

[tex]U = 73.20\ J[/tex]

Hence, The internal energy is 73.20 J.