Answer:
Velocity, V = 3t²- 28t+6
Displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = -249.05 m
v = -55.48 m/s
At t = 12.7 s
s = -141.48 m
v = 134.27 m/s
Explanation:
We have acceleration of a particle is given by a = 6t - 28
Velocity
[tex]v=\int adt=\int (6t - 28)dt=3t^2-28t+C[/tex]
At t = 0 we have v₀ = 6 m/s
v₀ = 6 = 3 x 0 ²-28 x 0+C
C = 6
So velocity, V = 3t² - 28t+6
Displacement
[tex]s=\int vdt=\int (3t^2-28t+6)dt=t^3-14t^2+6t+C[/tex]
At t = 0 we have s₀ = -8 m
s₀ = -8 = 0³ + 14 x 0² + 6 x 0 + C
C = -8
So displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m
v = 3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s
At t = 12.7 s
s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m
v = 3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s
A 2.00-kg block of aluminum at 50.0 °C is dropped into 5.00 kg of water at 20.0 °C. What is the change in entropy during the approach to equilibrium, assuming no heat is exchanged with the environment? The specific heat of aluminum is 0.22 cal/(g∙K).
To calculate the change in entropy during the approach to equilibrium, use the formula ΔS = mcΔT. For the aluminum block, ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum. For the water, ΔS_water = m_water * c_water * ΔT_water. Set ΔT_aluminum equal to ΔT_water and solve for T_water. Add the change in entropy for the aluminum and water to find the total change in entropy. Simplify the equation by converting cal to J and kg to g. Set the total change in entropy equal to zero and solve for T_aluminum to find the initial temperature of the aluminum.
Explanation:To calculate the change in entropy during the approach to equilibrium, we need to use the formula for entropy change:
ΔS = mcΔT
where ΔS is the change in entropy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the aluminum block is dropped into the water, so the final temperature will be the same for both the aluminum and the water. We can calculate the change in entropy for each substance separately and then add them together to find the total change in entropy.
For the aluminum block:
ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
ΔS_aluminum = 2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)
For the water:
ΔS_water = m_water * c_water * ΔT_water
ΔS_water = 5.00 kg * 1 cal/(g∙K) * (T_water - 20.0 °C)
Since the final temperature is the same for both substances, we can set ΔT_aluminum equal to ΔT_water and solve for T_water:
50.0 °C - T_aluminum = T_water - 20.0 °C
70.0 °C - T_aluminum = T_water
Substituting this value into the equation for ΔS_water:
ΔS_water = 5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C]
Now we can add the change in entropy for the aluminum and water:
ΔS_total = ΔS_aluminum + ΔS_water
ΔS_total = (2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
To simplify the equation, we can convert cal to J by multiplying by 4.184 and divide both sides by 1000 to convert kg to g:
ΔS_total = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
We can now solve for T_aluminum by setting ΔS_total equal to 0:
0 = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
Simplifying the equation:
0 = (8.368 J/(g∙K) * (50.0 °C - T_aluminum)) + (20.92 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))
Combining like terms:
0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))
Simplifying further:
0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum) - 104.6 J/(g∙K) * (70.0 °C - T_aluminum)
Expanding the equation:
0 = (29.288 J/(g∙K) * 50.0 °C - 29.288 J/(g∙K) * T_aluminum) - (104.6 J/(g∙K) * 70.0 °C - 104.6 J/(g∙K) * T_aluminum)
Simplifying the equation further:
0 = (1464.4 J - 29.288 J/(g∙K) * T_aluminum) - (7322 J - 104.6 J/(g∙K) * T_aluminum)
Combining like terms:
0 = 1464.4 J - 29.288 J/(g∙K) * T_aluminum - 7322 J + 104.6 J/(g∙K) * T_aluminum
Simplifying the equation:
0 = -5857.6 J + 75.312 J/(g∙K) * T_aluminum
Isolating T_aluminum:
5857.6 J = 75.312 J/(g∙K) * T_aluminum
T_aluminum = (5857.6 J) / (75.312 J/(g∙K))
T_aluminum ≈ 77.7 °C
Therefore, the initial temperature of the aluminum was approximately 77.7 °C.
Suppose there is a pendulum with length 5m hanging from a ceiling. A ball of mass 2kg is attached is attached to the bottom of the pendulum. The ball begins at rest. If I give the ball a velocity of 6 m/s, what is the maximum height that the ball will achieve? Use the energy conservation model to solve, and assume that there is no friction or air resistance.
Answer:
1.84 m from the initial point (3.16 m from the ceiling)
Explanation:
According to the law of conservation of energy, the initial kinetic energy of the ball will be converted into gravitational potential energy at the point of maximum height.
Therefore, we can write:
[tex]\frac{1}{2}mv^2 = mg\Delta h[/tex]
where
m = 2 kg is the mass of the ball
v = 6 m/s is the initial speed of the ball
g = 9.8 m/s^2 is the acceleration due to gravity
[tex]\Delta h[/tex] is the change in height of the ball
Solving for [tex]\Delta h[/tex],
[tex]\Delta h = \frac{v^2}{2g}=\frac{6^2}{2(9.8)}=1.84 m[/tex]
So, the ball raises 1.84 compared to its initial height.
Therefore:
- if we take the initial position of the ball as reference point, its maximum height is at 1.84 m
- if we take the ceiling as reference point, the maximum height of the ball will be
5 m - 1.84 m = 3.16 m from the ceiling
A motorcycle is moving at 18 m/s when its brakes are applied, bringing the cycle to rest in 4.7 s. To the nearest meter, how far does the motorcycle travel while coming to a stop?
Answer:
the motorcycle travels 42.4 meters until it stops.
Explanation:
Vi= 18 m/s
Vf= 0 m/s
t= 4.7 sec
Vf= Vi - a*t
deceleration:
a= Vi/t
a= 18m/s / 4.7 sec => a=-3.82 m/s²
x= Vi*t - (a * t²)/2
x= 42.4m
A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Answer:
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
Explanation:
Force per unit length between two current carrying wires is given by the formula
[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]
here we know that
[tex]F = 7.83 \times 10 N/m[/tex]
[tex]d = 7.0 cm = 0.07 m[/tex]
[tex]i_1 = 28 A[/tex]
now we will have
[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]
[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
Which of the following systems has constant kinetic and potential energies? A car moving along a level road at constant speed
A car moving up a hill at constant speed
A car moving down a hill at constant speed
All of the above
A uniform steel plate has an area of 0.819 m2. When subjected to a temperature difference between its sides, a heat current* of 31700 W is found to flow through it. What is the temperature gradient? What is the temperature difference when the plate is 0.0475 m thick? The thermal conductivity of steel is 50.2 W/(m·K).
Answer:
ΔT / Δx = 771 K/m
ΔT = 771 x 0.0475 = 36.62 k
Explanation:
P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k
Use the formula of conduction of heat
H / t = K A x ΔT / Δx
So, ΔT / Δx = P / K A
ΔT / Δx = 31700 / (50.2 x 0.819)
ΔT / Δx = 771 K/m
Now
ΔT = 771 x 0.0475 = 36.62 k
24. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?
Answer:
0.273 m/s
Explanation:
Momentum is conserved:
mu = (m + M) v
(0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v
v = 0.273 m/s
The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.The final velocity will be 0.273 m/sec.
What is law of conservation of momentum?The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.
The given data in the problem is;
m is the mass of clay model before collision= 0.200 Kg
u is the sliding velocity before collision=0.75 m/sec
M is the mass after collision= 0.350 Kg
v is the velocity after collision=?
On applying the law of conservation of momentum we get;
[tex]\rm mu = (m + M) v \\\\\ (0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v \\\\ v = 0.273 m/s[/tex]
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An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.5 cm, and the electric field within the capacitor has a magnitude of 2.5 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?
Answer:
6 x 10⁻¹⁵ J
Explanation:
d = distance between the plates = 1.5 cm = 0.015 m
E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m
q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C
Force on the electron due to electric field is given as
F = q E
F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)
F = 4 x 10⁻¹³ N
KE₀ = initial kinetic energy of electron at negative plate = 0 J
KE = final kinetic energy of electron at positive plate = ?
Using work-change in kinetic energy
F d = KE - KE₀
(4 x 10⁻¹³) (0.015) = KE - 0
KE = 6 x 10⁻¹⁵ J
An electron released from rest towards the positive plate of a capacitor, which is separated by a distance of 1.5 cm and an electric field of 2.5 x 10^6 V/m, attains a kinetic energy of 37.5 keV when it reaches the positive plate.
Explanation:The subject you're studying is called electrical potential energy, specifically its conversion into kinetic energy in the context of a parallel plate capacitor. In the case of an electron released from rest towards the positive plate of a capacitor, it is said to be moving through an electrical potential difference. This difference, coupled with the charge of the electron, provides the electron with energy, accelerating it.
Given that the electric field (E) is 2.5 x 106 V/m and the distance between the plates (d) is 1.5 cm or 0.015 m, we can use the formula E = V/d to calculate the potential difference (V). Substituting the given values, the potential difference is 2.5 x 106 V/m * 0.015 m = 37,500 V or 37.5 kV.
Furthermore, as per the relation that an electron accelerated through a potential difference of 1 V attains an energy of 1 electron-volt (eV), an acceleration through 37.5 kV will grant an energy of 37.5 keV. Since its initial kinetic energy was zero (as it was released from rest), this 37.5 keV is the kinetic energy of the electron just as it reaches the positive plate.
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To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
Answer:
4.4 square meters = 47 square foot
Explanation:
We have
1 meter = 3.28084 foot
1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot
4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot
4.4 square meters = 47 square foot
A parallel plate capacitor of area A = 30 cm2 and separation d = 5 mm is charged by a battery of 60-V. If the air between the plates is replaced by a dielectric of κ = 4 with the battery still connected, then what is the ratio of the initial charge on the plates divided by the final charge on the plates?
Answer:
0.25
Explanation:
A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²
d = separation between the plates = 5 mm = 5 x 10⁻³ m
[tex]C_{air}[/tex] = Capacitance of capacitor when there is air between the plates
k = dielectric constant = 4
[tex]C_{dielectric}[/tex] = Capacitance of capacitor when there is dielectric between the plates
Capacitance of capacitor when there is air between the plates is given as
[tex]C_{air} = \frac{\epsilon _{o}A}{d}[/tex] eq-1
Capacitance of capacitor when there is dielectric between the plates is given as
[tex]C_{dielectric} = \frac{k \epsilon _{o}A}{d}[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{\frac{\epsilon _{o}A}{d}}{\frac{k \epsilon _{o}A}{d}}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{k}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{4}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=0.25[/tex]
Charge stored in the capacitor when there is air is given as
[tex]Q_{air}=C_{air}V[/tex] eq-3
Charge stored in the capacitor when there is dielectric is given as
[tex]Q_{dielectric}=C_{dielectric}V[/tex] eq-4
Dividing eq-3 by eq-4
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}V}{C_{dielectric} V}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}}{C_{dielectric}}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=0.25[/tex]
If a copper wire has a resistance of 23.7 Ω at a temperature of 20.3 oC, what resistance does it have at 79.0 oC? (Neglect any change in length or cross-sectional area resulting from the change in temperature.)
Answer:
[tex]R_{79} = 28.91 OHM[/tex]
Explanation:
Resistance ca be determine by using following formula
[tex]R = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
where
R = Conductor resistance = 23.7Ω
Rref = conductor reistance at reference temperature,
α = temperature coefficient of resistance for material, for copper 40.41*10^{-4}
T = Conductor temperature in Celcius = 20.3°C
Tref = reference temperature at which α is specified.
[tex]23.7 = R_{ref}\left [ 1+ 40.41*10^{-4} (20.3) \right ][/tex][tex]R_{ref} =21.92 OHM[/tex]
now for 79 degree celcius
[tex]R_{79} = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
[tex]R_{79} =21.92\left [ 1+ 40.41*10^{-4} (79) \right ][/tex]
[tex]R_{79} = 28.91 OHM[/tex]
A tennis ball is thrown from a 25 m tall building with a zero initial velocity. At the same moment, another ball is thrown from the ground vertically upward with an initial velocity of 17 m/s. At which height will the two balls meet?
Answer:
The two balls meet in 1.47 sec.
Explanation:
Given that,
Height = 25 m
Initial velocity of ball= 0
Initial velocity of another ball = 17 m/s
We need to calculate the ball
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2+h[/tex]
Where, u = initial velocity
h = height
g = acceleration due to gravity
Put the value in the equation
For first ball
[tex]s_{1}=0-\dfrac{1}{2}gt^2+25[/tex]....(I)
For second ball
[tex]s_{2}=17t-\dfrac{1}{2}gt^2+0[/tex]....(II)
From equation (I) and (II)
[tex]-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0[/tex]
[tex]t=\dfrac{25}{17}[/tex]
[tex]t=1.47\ sec[/tex]
Hence, The two balls meet in 1.47 sec.
The two balls will meet at a height of approximately 16.465 meters above the ground.
To find the height at which the two balls meet, we need to consider their respective equations of motion and solve for the height when they intersect.
For the tennis ball thrown from the building:
[tex]\[ h_1(t) = 25 - \frac{1}{2} g t^2 \][/tex]
For the ball thrown from the ground:
[tex]\[ h_2(t) = 17t - \frac{1}{2} g t^2 \][/tex]
To find the time when they meet, we'll set:
[tex]\[ 25 - \frac{1}{2} g t^2 = 17t - \frac{1}{2} g t^2 \]\[ 25 = 17t \]\[ t = \frac{25}{17} \][/tex]
Now, substitute this value of \( t \) into either equation to find the height at which they meet:
[tex]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \left(\frac{25}{17}\right)^2 \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 9.8 \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 17.07 \]\[ h_1\left(\frac{25}{17}\right) = 25 - 8.535 \]\[ h_1\left(\frac{25}{17}\right) = 16.465 \][/tex]
A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and total charge q. What force does the wire experience?
Answer:
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
Explanation:
As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by
[tex]\lambda = \frac{q}{L}[/tex]
now the electric field due to long line charge at a distance d from it is given as
[tex]E = \frac{2k\lambda}{d}[/tex]
[tex]E = \frac{q}{2\pi \epsilon_0 d}[/tex]
now the force on the other charge in this electric field is given as
[tex]F = QE[/tex]
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
A falling baseball has an acceleration of magnitude 2.1 m/s2. What is its acceleration in feet per second squared?
Answer:
6.889 ft/s^2
Explanation:
1 m = 3.28 feet
So, 2.1 m/s^2 = 2.1 × 3.28 ft/s^2
= 6.889 ft/s^2
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
[tex]U=\dfrac{1}{2}kx^2[/tex]...(I)
Using gravitational potential energy
[tex]U' =mgh[/tex]....(II)
Using energy of conservation
[tex]E_{i}=E_{f}[/tex]
[tex]U_{i}+U'_{i}=U_{f}+U'_{f}[/tex]
[tex]\dfrac{1}{2}kx^2+0=0+mgh[/tex]
[tex]h=\dfrac{kx^2}{2mg}[/tex]
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
[tex]h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}[/tex]
[tex]h=11.653\ m[/tex]
Hence, The maximum height above the point of release is 11.653 m.
A small object of mass 3.66g and charge-19 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What ares the magnitude and direction of the electric field? magnitude N/C direction
Explanation:
It is given that,
Mass of the object, m = 3.66 kg
Charge, q = -19 μC = -19 × 10⁻⁶ C
It is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground such that,
[tex]F_g=F_e[/tex]
[tex]F_g\ and\ F_e[/tex] are gravitational and electrostatic forces respectively
[tex]mg=qE[/tex]
[tex]E=\dfrac{mg}{q}[/tex]
[tex]E=\dfrac{3.66\ kg\times 9.8\ m/s^2}{-19\times 10^{-6}\ C}[/tex]
E = −1887789.47 N/C
[tex]E=-1.89\times 10^6\ N/C[/tex]
Negative sign shows that the electric field is in the opposite direction of the electric force. Since, the weight of the object is in downward direction and its electric force (which is balancing its weight) is in upward direction. So, we can say that the electric field is in downward direction.
A sports car accelerates in third gear from 48.5 km/h to 80.2 km/h in 3.6 s. (a) What is the average acceleration of the car? (in m/s^2)
(b) If the car maintained this acceleration after reaching 80.2 km/h, how fast would it be moving 4.0 seconds later? (in km/h)
The average acceleration of the car is 2.45 m/s^2. If the car maintains this acceleration, it would be moving at a speed of 117.65 km/h 4.0 seconds later.
Explanation:(a) To find the average acceleration, we can use the formula: average acceleration (a) = (final velocity - initial velocity) / time. Converting the velocities to m/s gives us 13.47 m/s and 22.28 m/s, respectively. Plugging in the values, we get: a = (22.28 - 13.47) m/s / 3.6 s = 2.45 m/s^2. Therefore, the average acceleration of the car is 2.45 m/s^2.
(b) Since the car is maintaining the same acceleration, we can use the kinematic equation: final velocity = initial velocity + acceleration * time. Converting the initial velocity to m/s, we have 22.28 m/s. Plugging in the values, we get: final velocity = 22.28 m/s + 2.45 m/s^2 * 4.0 s = 32.68 m/s. Converting back to km/h gives us 32.68 m/s * 3.6 km/h/m = 117.65 km/h.
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Final answer:
The average acceleration of the sports car is approximately 2.446 m/s^2, and if it maintained this acceleration for an additional 4.0 seconds, it would be moving at roughly 115.4 km/h.
Explanation:
Calculating the Average Acceleration and Future Velocity
To find the average acceleration (a) of the car, we use the formula a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time taken for the change in velocity. First, we need to convert the velocities from km/h to m/s by multiplying by (1000 m/km) / (3600 s/h).
v_i = 48.5 km/h = (48.5 * 1000) / 3600 m/s ≈ 13.472 m/s
v_f = 80.2 km/h = (80.2 * 1000) / 3600 m/s ≈ 22.278 m/s
Now, we can calculate the average acceleration:
a = (22.278 m/s - 13.472 m/s) / 3.6 s ≈ 2.446 m/s²
To find the velocity 4.0 seconds after the car reaches 80.2 km/h, we use the formula v = v_f + a * t:
v = 22.278 m/s + 2.446 m/s² * 4.0 s ≈ 32.062 m/s
To convert this back to km/h:
v = 32.062 m/s * (3600 s/h) / (1000 m/km) ≈ 115.4 km/h
A single conducting loop of wire has an area of 7.26E-2 m2 and a resistance of 117 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.289 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.367 A?
Answer:
[tex]\frac{dB}{dt}[/tex] = 591.45 T/s
Explanation:
i = induced current in the loop = 0.367 A
R = Resistance of the loop = 117 Ω
E = Induced voltage
Induced voltage is given as
E = i R
E = (0.367) (117)
E = 42.939 volts
[tex]\frac{dB}{dt}[/tex] = rate of change of magnetic field
A = area of loop = 7.26 x 10⁻² m²
Induced emf is given as
[tex]E = A\frac{dB}{dt}[/tex]
[tex]42.939 = (7.26\times 10^{-2})\frac{dB}{dt}[/tex]
[tex]\frac{dB}{dt}[/tex] = 591.45 T/s
When the molecules in a body move with increased speed, it's possible that the body will change from a: A) gas to a solid B) gas to a liquid C) liquid to a solid. D) liquid to a gas.
Answer:
Liquid to a gas
Explanation:
When the molecules in a body move with increased speed, it's possible that the body will change from liquid to gas. The speed is increasing means they have more kinetic energy. The molecules of gas are very far apart from each other. They have much space between them so that they can move freely.
So, when the molecules move with increased speed, the body will change from liquid to gas. Hence, the correct option is (d) " liquid to gas".
In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit price in dollars, find the equilibrium quantity and the equilibrium price. p = −x2 − 3x + 80 and p = 7x + 5
In this exercise we have to use the knowledge of finance to calculate the equilibrium value and the quantity, so we have:
Equilibrium quantity = 5Equilibrium price = 40Organizing the information given in the statement we have:
p = -x²-3x+80p = 7x+5So equating the two given equations we have:
[tex]-x^2 - 3x + 80 = 7x+5\\x^2 +3x + 7x + 5 - 80 = 0\\x^2 + 10x - 75 = 0\\x^2- 5x + 15x -75 = 0\\x(x-5) + 15(x-5) = 0[/tex]
So we can see that the roots will be x = 5 and x = -15 since the quantity cannot be in negative therefore, the equilibrium quantity will be = 5 So replace that value at:
[tex]p = -(5)^2-3(5) + 80 = 40\\p = 7(5) + 5 = 40[/tex]
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A square, single-turn coil 0.132 m on a side is placed with its plane perpendicular to a constant magnetic field. An emf of 27.1 mV is induced in the winding while the area of the coil decreases at a rate of 0.0785 m2 /s. What is the magnitude of the magnetic field? Answer in units of T.
Answer:
0.35 T
Explanation:
Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s
Use the Faraday's law of electromagnetic induction
e = rate of change of magnetic flux
Let b be the strength of magnetic field.
e = dФ / dt
e = d ( B A) / dt
e = B x dA / dt
0.0271 = B x 0.0785
B = 0.35 T
The magnitude of the magnetic field is 3.45 × 10^-4 T.
Explanation:The magnitude of the induced emf in a square, single-turn coil can be calculated using the equation: emf = -N * A * (d(B)/dt), where N is the number of turns, A is the area of the coil, and d(B)/dt is the rate of change of the magnetic field. In this case, the area of the coil is decreasing at a rate of 0.0785 m2/s, and the emf is given as 27.1 mV. We can rearrange the equation to solve for the magnitude of the magnetic field (B):
B = -emf / (N * d(A)/dt) = -27.1 mV / (1 * 0.0785 m2/s) = -345.222 T/s = 3.45 × 10^-4 T/s.
Therefore, the magnitude of the magnetic field is 3.45 × 10^-4 T.
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The pressure of 4.20 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?
Answer:
6.30 L
Explanation:
P1 = P, V1 = 4.20 L, T1 = T
P2 = P/3, V2 = ?, T2 = T/2
Where, V2 be the final volume.
Use ideal gas equation
[tex]\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}[/tex]
[tex]V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}[/tex]
By substituting the values, we get
V2 = 6.30 L
A parachutist bails out and freely falls 50 m. Then the parachute opens, and thereafter she deceler- ates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?
Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.
A cable applies a vertical force to a crate with a mass of 70.0 kg. It first lifts the crate to a height of 12.0 m above the floor, and then lowers it back to the floor. What is the total work done by the force?
Answer:
Answer to the Question:
Explanation:
In this case, the total work done by the cable is zero, since in the aforementioned problem, the work depends only on the starting and ending point, these two being equal. Keeping its gravitational potential energy equal.
Answer:
The total work done is zero.
Explanation:
Given;
mass of the crate, m = 70.0 kg
height above ground through which the crate is lifted, h = 12.0 m
The only work associated in lifting and lowering this crate is gravitational potential energy.
Potential Energy during lifting = - mgh
= - 70 x 9.8 x 12
= - 8232 J
Potential Energy during lowering = mgh
= 70 x 9.8 x 12
= 8232 J
Total total work done by the force = - 8232 J + 8232 J = 0
Therefore, the total work done is zero.
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C).1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.34 kg2.34 kg of this oil from 23 °C23 °C to 191 °C?191 °C?
Answer:
Heat energy required = 687.96 kJ
Explanation:
Heat energy required, H = mCΔT.
Mass of cooking oil, m = 2.34 kg = 2340 g
Specific heat of cooking oil, C = 1.75 J/(g⋅°C)
Initial temperature = 23 °C
Final temperature = 191 °C
Change in temperature, ΔT = 191 - 23 = 168 °C
Substituting values
H = mCΔT
H = 2340 x 1.75 x 168 = 687960 J = 687.96 kJ
Heat energy required = 687.96 kJ
A diver can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position. If she makes two rotations in 1.5 seconds when in the tuck position, what is her angular speed (rad/sec) when in the straight position?
Answer:
29.3 rad/s
Explanation:
Moment of inertia in straight position, I1 = I
He takes 2 rotations in 1.5 second
So, Time period, T = 1.5 / 2 = 0.75 second
w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s
Moment of inertia in tucked position, I2 = I / 3.5
Let the new angular speed is w2.
By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.
L1 = L2
I1 w1 = I2 w2
I x 8.37 = I / 3.5 x w2
w2 = 29.3 rad/s
To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. The moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. By applying the conservation of angular momentum equation, we can calculate the angular speed in the straight position.
Explanation:To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. According to the conservation of angular momentum, the product of the initial moment of inertia and initial angular speed is equal to the product of the final moment of inertia and final angular speed.
Let's assume the initial angular speed in the tuck position is denoted by w' and the final angular speed in the straight position is denoted by w. We are given that the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. Therefore, the final moment of inertia (I) is 3.5 times greater than the initial moment of inertia (I').
The conservation of angular momentum equation can be written as:
I' * w' = I * w
Since the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position, we have:
w = w' * (I' / I)
Using the given values, we can calculate the angular speed in the straight position.
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A long solenoid has a length of 0.67 m and contains 1700 turns of wire. There is a current of 5.5 A in the wire. What is the magnitude of the magnetic field within the solenoid?
B = 17.5mT.
A solenoid is a coil formed by a wire (usually copper) wound into a cylindrical spiral shape capable of creating a magnetic field that is extremely uniform and intense inside, and very weak outside.
To calculate the magnetic field generated inside the solenoid through which a current flows is given by the equation:
B = μ₀nI
Where μ₀ is the constant of magnetic proportionality of the vacuum (4π x 10⁻⁷T.m/A), n is the relation between the number of turns of wire and its length given by N/L and I is the current flowing through the solenoid.
Given a long solenoid of length 0.67m, 1700.00 turns of wire and a current flowing through the wire of 5.50A. Calculate the magnetic field inside the solenoid.
B = (4π x 10⁻⁷T.m/A)(1700turns/0.67m)(5.50A)
B = 0.0175T
B = 17.5mT
A superconducting solenoid is to be designed to have an interior field of 7.171 T with a maximum current of 1000 A. The permeability of free space is 1.25664 × 10−6 T · m/A. How many windings are required on a 1- meter solenoid length?
Answer:
5706.5
Explanation:
B = 7.171 T, i = 1000 A, μ0 = 1.25664 x 10^-6 T m/A, l = 1 m
Let the number of turns be N.
The magnetic field due to a current carrying solenoid is given by
B = μ0 x N x i / l
B x l / (μ0 x i) = N
N = 7.171 x 1 / (1.25664 x 10^-6 x 1000)
N = 5706.5
To design a superconducting solenoid that generates an interior field of 7.171 T with a maximum current of 1000 A, approximately 5709 turns per meter are required on a 1-meter length of the solenoid.
Explanation:To determine the number of windings required on a 1-meter solenoid to achieve a magnetic field of 7.171 T with a maximum current of 1000 A, we can use the formula for the magnetic field inside a solenoid, which is given by:
B = μ_0 * n * I
Where:
B is the magnetic field inside the solenoidμ_0 is the permeability of free space (μ_0 = 1.25664 × 10⁻⁶ T · m/A)n is the number of turns per meterI is the current through the solenoidRearranging the formula to solve for n, we get:
n = B / (μ_0 * I)
Substituting the given values:
n = 7.171 T / (1.25664 × 10⁻⁶ T · m/A × 1000 A)
Now, calculating the value:
n = 7.171 / (1.25664 × 10⁻⁶ × 1000)
n = 7.171 / (1.25664 × 10⁻³)
n = 5708.57 turns per meter
To achieve the desired magnetic field, the solenoid must have approximately 5709 turns per meter to accommodate the rounding to the nearest whole number, since we can't have a fraction of a turn.
In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel?
Answer:
18.7 m
Explanation:
[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s
θ = angle of launch from the horizontal = 47 deg
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 1.80 m
t = time of travel
Using the kinematics equation
[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]
- 1.80 = (9.5) t + (0.5) (- 9.8) t²
t = 2.11 s
Consider the motion along the horizontal direction
x = horizontal displacement of the shot
[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.11 s
Using the kinematics equation
[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]
x = (8.87) (2.11) + (0.5) (0) (2.11)²
x = 18.7 m
The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
What is projectile motion?A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.
Given to us
The initial velocity of the shotput, u = 13 m/s
The angle from the horizontal, θ = 47.0°
The height from which shotput was thrown h = 1.80 m
As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,
[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]
Substitute the values,
[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]
R = 18.7 m
Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
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An object moves uniformly around a circular path of radius 23.5 cm, making one complete revolution every 1.95 s. (a) What is the translational speed of the object? (b) What is the frequency of motion in hertz? (c) What is the angular speed of the object?
Explanation:
a)
The circumference of the path is:
C = 2πr
C = 2π (0.235 m)
C = 1.48 m
Velocity = displacement / time
v = 1.48 m / 1.95 s
v = 0.757 m/s
b)
1 rev / 1.95 s = 0.513 Hz
c)
1 rev / 1.95 s × (2π rad / rev) = 3.22 rad/s
(a) The translational speed of the object is 0.76 m/s.
(b) The frequency of the object's motion is 0.51 Hz.
(c) The angular speed of the object is 3.22 rad/s.
Angular speed of the objectThe angular speed of the object is calculated as follows;
ω = 1 rev/ 1.95 s = 0.51 rev/s
ω = 0.51 rev/s x 2π rad
ω = 3.22 rad/s
Angular frequency of the objectThe frequency of the object's motion is determined from the angular speed as shown below;
ω = 2πf
f = ω/2π
f = (3.22)/2π
f = 0.51 Hz
Translational speed of the objectThe translational speed of the object is calculated as follows;
v = ωr
v = 3.22 x 0.235
v = 0.76 m/s
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