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.
If T is the absolute temperature, the intensity of radiation from an ideal radiator will be proportional to (a) T (b) T (c) T (d) T (e) none of the above
Answer:
T^4
Explanation:
According to the Stefan's law, the energy radiated per unit time and per unit area is directly proportional to the four power of teh absolute temperature of the black body.
E ∝ T^4
The intensity of radiation from an ideal radiator is not directly proportional to the absolute temperature T, but rather to T4, according to the Stefan-Boltzmann law of radiation.
Explanation:The intensity of radiation from an ideal radiator, or blackbody, is not directly proportional to the absolute temperature T; instead, it follows the Stefan-Boltzmann law of radiation. This law states that the intensity of radiation, referred to as 'P' or power, is directly proportional to the fourth power of the absolute temperature (P = σAT4). Here, σ (sigma) represents Stefan-Boltzmann constant, A denotes surface area, and T signifies absolute temperature.
This means the answer to your question is none of the above, as the intensity of the radiation isn't directly proportional to T, but rather to T4.
As an example, considering two temperatures T₁ = 293 K and T₂ = 313 K, the rate of heat transfer with T₂ is about 30 percent higher than T₁. This showcases the considerable impact of temperature on radiation intensity.
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A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Answer:
3.95 m
Explanation:
m = 1 kg, h = 100 m, k = 125 N/m
Let the spring is compressed by y.
Use the conservation of energy
potential energy of the mass is equal to the energy stored in the spring
m x g x h = 1/2 x ky^2
1 x 9.8 x 100 = 0.5 x 125 x y^2
y^2 = 15.68
y = 3.95 m
Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilometers will the train have traveled after 4.00 hours? Round to the nearest km.
Answer:
605 km
Explanation:
Hello
the same units of measure should be used, then
Step 1
convert 42 m/s ⇒ km/h
1 km =1000 m
1 h = 36000 sec
[tex]42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\[/tex]
[tex]0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}[/tex]
Step 2
find kilometers traveled after 4 hours
[tex]V=\frac{s}{t}\\ \\[/tex]
V,velocity
s, distance traveled
t. time
now, isolating s
[tex]V=\frac{s}{t} \\s=V * t\\[/tex]
and replacing
[tex]s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\[/tex]
S=604.8 Km
Have a great day
Work is required to compress 5.00 mol of air at 20.00C and 1.00 atm to one-tenth of the original volume by an adiabatic process. How much work is required to produce this same compression?
Answer:46.03 KJ
Explanation:
no of moles(n)=5
temperature of air(T)=[tex]20^{\circ}[/tex]
pressure(p)=1 atm
final volume is [tex]\frac{V}{10}[/tex]
We know work done in adaibatic process is given by
W=[tex]\frac{P_iV_i-P_fV_f}{\gamma -1}[/tex]
[tex]\gamma[/tex] for air is 1.4
we know [tex]P_iV_i^{\gamma }= P_fV_f^{\gamma }[/tex]
[tex]1\left ( V\right )^{\gamma }[/tex]=[tex]P_f\left (\frac{v}{10}\right )^{\gamma }[/tex]
[tex]10^{\gamma }=P_f[/tex]
[tex]P_f=25.118 atm[/tex]
W=[tex]\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}[/tex]
W=-46.03 KJ
it means work is done on the system
If the charge of a particle doubles, what happens to the force acting on it?
It doubles
It gets reduced by a factor of two
It stays the same
A charge exerts a negative force on another charge. Does that mean that:
Both charges are positive
Both charges are negative
the charges are of opposite signs
please explain this throughly! thanks
Explanation:
(1) The force that exists between charged particles is electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Where
q₁ and q₂ are charges
r is distance between charges
If the charge of a particle doubles, the electric force doubles. So, the correct option is (a) "It doubles".
(2) A charge exerts a negative force on another charge. Negative force denotes the force is attractive. It means that the charges are of opposite sign. So, the correct option is (c) "the charges are of opposite signs".
The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary to increase the object's temperature from 6°C to 12°C? 4Q 2Q Q 3Q
Answer:
The answer is 3Q
Explanation:
The metal temperature increases in a linear way, we could get a difference between final and initial temperature
[tex]DT=FinalTemperature-InitialTemperature[/tex]
We get a temperature difference of 2 degrees per each heat addition. If we add the same heat 3 times more, it will increase to 12 degrees
Answer:
The quantity of heat required to increase the temperature of the object from 6°C to 12°C is 3Q
Explanation:
Heat capacity is the quantity of heat required to increase the temperature of an object.
Q = mcΔθ
where;
Q is the quantity of heat
m is the mass of the object
c is specific heat capacity of the object
Δθ is change in temperature = T₂ - T₁
For the first sentence of this question;
Q = mc(6-4)
Q = mc(2)
Q = 2mc
For the second sentence of this question;
Let Q₂ be the quantity of heat required to increase the temperature of the object from 6°C to 12°C
Q₂ = mcΔθ
Q₂ = mc(12-6)
Q₂ = mc(6)
Q₂ = 6mc
Q₂ = 3(2mc)
Recall, Q = 2mc
Thus, Q₂ = 3Q
A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
Answer:
a) Maximum height reached above ground = 2.8 m
b) When he reaches maximum height he is 2 m far from end of the ramp.
Explanation:
a) We have equation of motion v²=u²+2as
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0²=5.6² + 2 x -9.81 x s
s = 1.60 m
Height above ground = 1.2 + 1.6 = 2.8 m
b) We have equation of motion v= u+at
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0= 5.6 - 9.81 x t
t = 0.57s
Now considering horizontal motion of skateboarder.
We have equation of motion s =ut + 0.5 at²
u = 6.6cos58 = 3.50 m/s
a = 0 m/s²
t = 0.57
Substituting
s =3.5 x 0.57 + 0.5 x 0 x 0.57²
s = 2 m
When he reaches maximum height he is 2 m far from end of the ramp.
The highest point reached by the skateboarder is 1.612 meters above the ground. When the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is 3.568 meters.
Given:
Initial velocity (v₀) = 6.6 m/s
Launch angle (θ) = 58°
Height of the ramp (h) = 1.2 m
Acceleration due to gravity (g) = 9.8 m/s²
(a) To find the maximum height reached by the skateboarder:
Δy = v₀y² / (2g)
v₀y = v₀ × sin(θ)
v₀y = 6.6 × sin(58°)
v₀y = 5.643 m/s
Δy = (5.643 )² / (2 × 9.8)
Δy ≈ 1.612 m
Therefore, the highest point reached by the skateboarder is 1.612 meters above the ground.
(b) The time of flight can be calculated using the equation:
t = 2 × v₀y / g
t = 2 × 5.643 / 9.8
t = 1.153 s
Δx = v₀x × t
First, we need to find the initial horizontal velocity (v₀x):
v₀x = v₀ × cos(θ)
v₀x = 6.6 × cos(58°)
v₀x = 3.099 m/s
Δx = 3.099 * 1.153
Δx = 3.568 m
Therefore, when the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is 3.568 meters.
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A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30°. The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?
Answer:
40N
Explanation:
frictional force = mgsin©
frictional force = 8*10*sin30
frictional force = 8*10*0.5= 40N//
Sound 1 has an intensity of 45.0 W/m2. Sound 2 has an intensity level that is 3.2 dB greater than the intensity level of sound 1. What is the intensity level of sound 2?
Answer:
The intensity level of sound 2 is 93.3\ W/m².
Explanation:
Given that,
Intensity of sound 1 = 45.0 W/m²
Intensity level of sound 2 = 3.2 dB
[tex]I_{0}=10^{-12}\ w/m^2[/tex]
We need to calculate the intensity
Using equation of the sound level intensity
[tex]I_{dB}=10 log\dfrac{I}{I_{0}}[/tex]
[tex]I_{dB}=10 log(\dfrac{45.0}{10^{-12}})[/tex]
[tex]I_{dB}=136.5 dB[/tex]
The intensity of sound 2 is greater than 3.2 dB.
Therefore,
[tex]I_{2}=136.5+3.2[/tex]
[tex]I_{2}=139.7\ dB[/tex]
Calculate the intensity of sound 2
[tex]I_{2}=10log(\dfrac{I'}{10^{-12}})[/tex]
[tex]I'=10^{-12}(10^{\frac{139.7}{10}})[/tex]
[tex]I'=93.3\ W/m^2[/tex]
Hence, The intensity level of sound 2 is 93.3\ W/m².
Decibels, measured in dB, quantifies the intensity level of sound. They relate to the intensity of sound in a logarithmic manner. More information and advanced calculations are needed to find the actual intensity of Sound 2 based on the given decibel difference
Explanation:Let's begin by understanding that decibels (dB) measure the intensity level of sound. The decibel level (dB) is a logarithmic unit used to describe a ratio of two values of a physical quantity, here, the sound intensity. One key fact is that each time the intensity increases by a factor of 10; the decibel level rises by 10 dB. Thus, the difference in decibels corresponds to the logarithmic ratio of the two intensities involved.
In the given problem, Sound 2 is 3.2 dB greater than Sound 1. It means the ratio of their intensities has a logarithmic relation with a value of 3.2. However, to convert this decibel difference into an actual value for Sound 2's intensity, additional calculations involving logarithms and the initial intensity value for Sound 1 are needed. As these calculations involve advanced math, it'd be more appropriate to take this problem to a higher level course or consult a logarithm table or a scientific calculator.
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Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to Lonnie with a return speed of 60 m/s. If the bat is in contact with the ball for 0.050 s what is the impulse experienced by the ball? A. 360 N.s B. 20 N.s C. 400 N.s D. 9.0 N.s
Answer:
B) 20N.s is the correct answer
Explanation:
The formula for the impulse is given as:
Impulse = change in momentum
Impulse = mass × change in speed
Impulse = m × ΔV
Given:
initial speed = 40m/s
Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)
mass = 0.20 kg
Thus, we have
Impulse = 0.20 × (40m/s - (-60)m/s)
Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s
The impulse experienced by the ball is,
[tex]\rm Impulse = 20\; Nsec[/tex]
Given :
Mass = 0.20 Kg
Initial Speed = 40 m/sec
Final Speed = -60 m/sec (minus sign shows that ball move in opposite direction)
Solution :
We know that the formula of Impulse is,
[tex]\rm Impulse = m\times \Delta V[/tex]
where, m is the mass and [tex]\rm \Delta V[/tex] is change in velocity.
[tex]\rm Impulse = 0.20\times(40-(-60))[/tex]
[tex]\rm Impulse = 20\; Nsec[/tex]
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An object with mass 80 kg moved in outer space. When it was at location <7, -34, -7> its speed was 14.0 m/s. A single constant force <200, 460, -150> N acted on the object while the object moved to location <12, -42, -11> m. What is the speed of the object at this final location?
The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.
What is the speed of a body?The speed of a body is the rate at which it covers the total distance is in the time taken.
If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.
The constant force applied on the object is <200, 460, -150> N. It can be written in the vector form as,
[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]
The initial position vector of the object for the location <7, -34, -7> can be given as,
[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]
The final position vector of the object is for the location <12, -42, -11> m can be given as,
[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]
Thus, the total distance traveled by the constant force is,
[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]
The work done by a object is the product of force and displacement. Thus, work done on the object is,
[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]
Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,
[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]
Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,
[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]
Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.
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To find the speed of the object at the final location, we need to calculate the net force acting on it using the equation F = ma. By plugging in the values, we can calculate the final speed of the object.
Explanation:To determine the speed of the object at the final location, we need to calculate the net force acting on the object using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.
Given that the force acting on the object is <200, 460, -150> N, and the mass of the object is 80 kg, we can calculate the acceleration using the formula a = F/m. After calculating the acceleration, we can use it to find the final speed of the object using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement.
By plugging in the given values, we can calculate the final speed of the object at the location <12, -42, -11> m.
A car of mass 2800 kg collides with a truck of mass 4000 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was <36, 0, 0> m/s, and the truck's velocity just before the collision was <-15, 0, 29> m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)?
(a) The velocity of the car-truck system after collision is 18.08 m/s.
(b) The increase in internal energy of the car and truck is 2,835,031.2 J.
Velocity of the truck - car system after collision
The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;
Final velocity in x - directionm₁u₁ + m₂u₂ = v(m₁ + m₂)
2800(36) + 4000(-15) = vx(2800 + 4000)
40,800 = 6800vx
vx = 6 m/s
Final velocity in z - directionm₁u₁ + m₂u₂ = v(m₁ + m₂)
2800(0) + 4000(29) = vz(2800 + 4000)
116,000 = 6800vz
vz = 17.06 m/s
Resultant velocity of the car-truck system after the collision [tex]v= \sqrt{v_x^2 + v_z^2} \\\\v = \sqrt{6^2 + 17.06^2} \\\\v = 18.08 \ m/s[/tex]Initial kinetic energy of the car and truckK.E(car) = ¹/₂mv²
K.E(car) = ¹/₂ x (2800) x (36)²
K.E(car) = 1,814,400 J
v(truck) = √(15² + 29²) = 32.65
K.E(truck) = ¹/₂ x (4000) x (32.65)²
K.E(truck) = 2,132,045 J
K.E(total) = 1,814,400 J + 2,132,045 J = 3,946,445 J
Final kinetic energy of the systemK.E = ¹/₂(m₁ + m₂)v²
K.E = ¹/₂ x (2800 + 4000) x (18.08)²
K.E = 1,111,413.8 J
Increase in internal energyU = ΔK.E
U = 3,946,445 J - 1,111,413.8 J
U = 2,835,031.2 J
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Final answer:
To determine the velocity of the stuck-together car and truck after the collision, apply the principle of conservation of momentum. Consider the change in kinetic energy and the production of thermal energy and deformation.
Explanation:
To determine the velocity of the stuck-together car and truck just after the collision, we can apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. We can find the final velocity by summing the momenta of the car and truck and dividing by their combined mass.
In this case, the car's momentum is the product of its mass and velocity, and the truck's momentum is the product of its mass and velocity. Adding these momenta together and dividing by the combined mass gives us the final velocity of the stuck-together car and truck just after the collision.
For the increase in internal energy of the car and truck, we need to consider the change in kinetic energy and the production of thermal energy and deformation. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energies of the car and truck. The thermal energy and deformation depend on factors such as the materials involved and the severity of the collision.
You are in a fire truck heading directly toward a tunnel in a mountainside at 17m/s. you hear the truck's 90Hz siren as well as its reflection from the mountainside. What beat frequency do you hear?
Hey there!:
Take the speed of sound to be 343m/s.
Direct frequency perceived by observer:
(343 + 17) / 343) * 90Hz = 94.460Hz
Change in frequency = ( 4.460 - 90 ) = 4.460Hz.
90 - (4.460 x 2) = 81.08 Hz. indirect frequency heard by observer.
Therefore :
Beat frequency = (94.460 - 81.08) = 13.38Hz.
Hope this helps!
A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end. What is the linear speed of the sphere when it reaches the bottom of the ramp
The Linear speed of the sphere is mathematically given as
v = 4.1 m/s
What is the linear speed of the sphere when it reaches the bottom of the ramp?Question Parameter(s):
A solid sphere of mass 4.0 kg and radius of 0.12 m starts from rest at the top of a ramp inclined 15°
Generally, the equation for the conservation of energy is mathematically given as
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2[/tex]
Therefore
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})\\\\mgh = \frac{7}{10}mv^2[/tex]
In conclusion, the Speed is
[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]
v = 4.1 m/s
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The linear speed of the sphere when it reaches the bottom of the ramp is approximately[tex]\( 4.85 \, \text{m/s} \).[/tex]
At the top of the ramp, the potential energy U of the sphere is given by:
[tex]\[ U = mgh \][/tex]
where m is the mass of the sphere, g is the acceleration due to gravity (approximately[tex]\( 9.8 \, \text{m/s}^2[/tex], and h is the height of the ramp.
At the bottom of the ramp, the kinetic energy K of the sphere is given by:
[tex]\[ K = \frac{1}{2}mv^2 \][/tex]
where v is the linear speed of the sphere.
Since energy is conserved, we have:
U = K
[tex]\[ mgh = \frac{1}{2}mv^2 \][/tex]
We can solve for v by canceling the mass m from both sides and multiplying through by 2:
[tex]\[ 2gh = v^2 \][/tex]
Taking the square root of both sides gives us the linear speed v :
[tex]\[ v = \sqrt{2gh} \][/tex]
Given that the height h is 1.2 m and the acceleration due to gravity g is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex], we can plug in these values:
[tex]\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.2 \, \text{m}} \][/tex]
[tex]\[ v = \sqrt{23.52 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v \ =4.85 \, \text{m/s} \][/tex]
Gina perceives the car to be far away because the sides of the road upon which it is moving seem to come together to be no wider than the car itself. This is an example of the ____ cue to depth.
Answer:
Linear Perspective.
Explanation:
Linear perspective cue to depth and it is in respect to both texture gradient or the next depth cue and relative size. In linear perspective, parallel lines that go along seems to have converged at distance very far, eventually reaching a vanishing point at the horizon. For example the railway track appears to be converging at a point very distant from the observer but it didn't.
Answer:
d. linear perspective
Explanation:
According to a different source, these are the options that come with this question:
a. visual acuity
b. texture gradient
c. retinal disparity
d. linear perspective
Linear perspective refers to a way in which depth is created. This term is particularly employed by artists who create an illusion of depth on a flat surface. This perspective leads all parallel lines to be lost in a single vanishing point on the horizon line. In this example, Gina is using the linear perspective cue to understand the depth of the scene she is observing.
When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt? (a) How much torque are you exerting in newtons x meters (relative to the center of the bolt) (b) Convert this torque to footpounds
Answer:
Part a)
[tex]\tau = 23.1 Nm[/tex]
Part b)
[tex]\tau = 17.05 Foot pound force[/tex]
Explanation:
As we know that torque is defined as the product of force and its perpendicular distance from reference point
so here we have
[tex]\tau = \vec r \times \vec F[/tex]
now we have
[tex]\tau = (0.140)(165)[/tex]
[tex]\tau = 23.1 Nm[/tex]
Part b)
Now we know the conversion as
1 meter = 3.28 foot
1 N = 0.225 Lb force
now we have
[tex]\tau = 23.1 Nm[/tex]
[tex]\tau = 23.1 (0.225 Lb)(3.28 foot)[/tex]
[tex]\tau = 17.05 Foot pound force[/tex]
The torque exerted when tightening a bolt with a force of 165 N at a distance of 0.140 m from the center of the bolt is 23.1 N·m. This torque can be converted to approximately 17.0 ft·lb.
Explanation:To find the torque exerted when tightening a bolt, you can use the equation T = RF sin(θ). In this case, the force is 165 N and the distance from the center of the bolt is 0.140 m. Since you are pushing perpendicularly, the angle θ is 90 degrees. Thus, the torque exerted is 165 N x 0.140 m x sin(90°) = 23.1 N·m.
To convert this torque to foot-pounds, you can use the conversion factor 1 N·m = 0.73756 ft·lb. Therefore, the torque is 23.1 N·m x 0.73756 ft·lb/N·m ≈ 17.0 ft·lb.
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A 1 cm^3 block with a density of 0.92 g/cm^3 is floating in a container of water (d =1g/ cm^3). You may ignore any air pressure throughout this problem. What buoyant force is necessary to keep the block from sinking?
Answer:
980 dyne
Explanation:
Volume = 1 cm^3, d = 0.92 g / cm^3, D = 1 g/cm^3
In the equilibrium condition, the buoyant force is equal to the weight of the block.
Buoyant force = Volume of block x density of water x g
Buoyant force = 1 x 1 x 980 = 980 dyne
You heat a 541cm^3 sample of a substance from 133°C to 273°C and find that its volume increases by 2.25 cm^3. Calculate the coefficient of volume expansion of this substance.
Answer:
[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]
Explanation:
As we know by the theory of expansion the change in the volume of the object is directly proportional to change in temperature and initial volume.
So here we can say
[tex]\Delta V = V_0\gamma \Delta T[/tex]
here
[tex]\gamma[/tex] = coefficient of volume expansion
so we have
[tex]\gamma = \frac{\Delta V}{V_0 \Delta T}[/tex]
now plug in all values
[tex]\gamma = \frac{2.25 cm^3}{(541 cm^3)(273 - 133)}[/tex]
[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]
A ball is dropped from rest. What will be its speed when it hits the ground in each case. a. It is dropped from 0.5 meter above the ground. b. It is dropped from 5 meters above the ground. c. It is dropped from 10 feet above the ground.
Answer:
(a) 3.13 m/s
(b) 9.9 m/s
(c) 7.73 m/s
Explanation:
u = 0 m/s, g = 9.8 m/s^2
Let v be the velocity of ball as it hit the ground.
(a) h = 0.5 m
Use third equation of motion.
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 9.8 x 0.5
v^2 = 9.8
v = 3.13 m/s
(b) h = 5 m
Use third equation of motion.
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 9.8 x 5
v^2 = 98
v = 9.9 m/s
(c) h = 10 feet = 3.048 m
Use third equation of motion.
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 9.8 x 3.048
v^2 = 59.74
v = 7.73 m/s
A vertically mounted spring (k = 750 N/m) is compressed by 35.0 cm relative to its unstrained length. A mass (m = 0.36 kg) is placed at rest against the spring. When the spring is released, the mass is launched vertically in the air. How high from release point the mass can reach?
Answer:
Height, h = 13.02 meters
Explanation:
It is given that,
Spring constant of the spring, k = 750 N/m
It is compressed by 35 cm relative to its unstained length, x = 35 cm = 0.35 m
Mass of the object, m = 0.36 kg
When the spring is released, the mass is launched vertically in the air. We need to find the height attained by the mass at this position. On applying the conservation of energy as :
Energy stored in the spring = change on potential energy
[tex]\dfrac{1}{2}kx^2=mgh[/tex]
[tex]h=\dfrac{kx^2}{2mg}[/tex]
[tex]h=\dfrac{750\ N/m\times (0.35\ m)^2}{2\times 0.36\ kg\times 9.8\ m/s^2}[/tex]
h = 13.02 meters
So, the mass will reach a height of 13.02 meters. Hence, this is the required solution.
The maximum height reached by the mass when the spring is released is 13.02 m.
The given parameters;
spring constant, k = 750 N/mextension of the spring, x = 35 cm = 0.35 mmass attached, m = 0.36 kgApply the principle of conservation energy to determine the maximum height reached by the mass when the spring is released.
mgh = ¹/₂kx²
where;
h is the maximum height reached by the massg is acceleration due to gravity[tex]h = \frac{kx^2}{2mg} \\\\h = \frac{(750) \times (0.35)^2}{2(0.36\times 9.8)} \\\\h = 13.02 \ m[/tex]
Thus, the maximum height reached by the mass when the spring is released is 13.02 m.
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Two banked curves have the same radius. Curve A is banked at 12.7 °, and curve B is banked at an angle of 15.1°. A car can travel around curve A without relying on friction at a speed of 19.1 m/s. At what speed can this car travel around curve B without relying on friction?
The speed of the car on Curve B is obtained by solving for the radius in Curve A's equation, then substituting that into the formula for Curve B. The formula involved is based on principles of circular motion and forces.
Explanation:The subject of this question is Physics, specifically the principles of circular motion and the forces at play within. Given the information from Curve A, we can ascertain that the speed of the car on Curve B can be found using principles of physics. The formula to find the speed at which the car can travel around a banked curve without relying on friction is: v = sqrt[rgtan(Θ)]. Where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and Θ is the angle of the banked curve.
Since the radius is the same for both curves, and the speed is known for Curve A, we can set it up so: 19.1 = sqrt[r * 9.8 * tan(12.7)]. We then solve for r (the radius), and apply it to Curve B: v = sqrt[r * 9.8 * tan(15.1)]. By substituting in the value of r obtained from the first equation, we can calculate the speed for Curve B.
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To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.
Explanation:The speed at which a car can travel around a banked curve without relying on friction can be calculated using the ideal banking angle formula. The formula is given by v = √(g * r * tan(θ)), where v is the speed, g is the acceleration due to gravity, r is the radius of the curve, and θ is the banked angle. In this case, curve A is banked at 12.7° and the car can travel at a speed of 19.1 m/s. To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.
Plugging in the values into the formula, v = √(9.8 * r * tan(15.1)). Since the radius is the same for both curves, we can solve for v: 19.1 = √(9.8 * r * tan(12.7))
Solving the equation for v, we find that the car can travel around curve B without relying on friction at a speed of approximately 21.2 m/s.
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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the magnitude of the electric flux througha rectangular area of 1.65 m2 in the xy-plane. N m2/C
Answer:
The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]
Explanation:
Given that,
Electric field = 2.35 V/m
Angle = 25.0°
Area [tex]A= 1.65 m^2[/tex]
We need to calculate the flux
Using formula of the magnetic flux
[tex]\phi=E\cdot A[/tex]
[tex]\phi = EA\cos\theta[/tex]
Where,
A = area
E = electric field
Put the value into the formula
[tex]\phi=2.35\times1.65\times\cos 25^{\circ}[/tex]
[tex]\phi=2.35\times1.65\times0.91[/tex]
[tex]\phi=3.53\ N-m^2/C[/tex]
Hence, The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]
In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of the wind and the speed of the airplane in still air. g
Answer:
Velocity of airplane is 500 km/h
Velocity of wind is 40 km/h
Explanation:
[tex]V_a[/tex]= Velocity of airplane in still air
[tex]V_w[/tex]= Velocity of wind
Time taken by plane to travel 1150 km against the wind is 2.5 hours
[tex]V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)[/tex]
Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours
[tex]V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)[/tex]
Subtracting the two equations we get
[tex]V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h[/tex]
Applying the value of velocity of wind to the first equation
[tex]V_a-40=460\\\Rightarrow V_a =500\ km/h[/tex]
∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h
The given mathematical problem involves calculating speeds. The speed of the airplane in still air was found to be 501 km/h and the wind speed was calculated to be 41 km/h, by using the concept of speed equals distance divided by time.
Explanation:The subject is related to measurements of speed, which is essentially a mathematical problem, specifically involving algebraic calculations. In order to solve this problem, we will use the concept that speed equals distance divided by time.
Firstly, convert all time measurements to the same unit. Since the speed of an airplane is typically measured in km/h, it will be convenient to convert the 50 minutes to hours (50/60 = 0.83 hours).
When moving against the wind, the combined speed (airplane speed minus wind speed) is 1150 km / 2.5 h = 460 km/h. When moving with the wind, the combined speed (airplane speed plus wind speed) becomes 450 km / 0.83 h = 542 km/h.
Now you can find the speed of the airplane in still air by averaging the two combined speeds: (460 km/h + 542 km/h) / 2 = 501 km/h. The speed of the wind can be found by subtracting the airplane's speed from the higher combined speed: 542 km/h - 501 km/h = 41 km/h. Therefore, the speed of the airplane in still air is 501 km/h and the wind speed is 41 km/h.
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A rigid tank contains 1.80 moles of helium, which can be treated as an ideal gas, at a pressure of 27.3 atm. While the tank and gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.20 atm. How many moles are removed?
Answer:
Has been removed 1.458 moles.
Explanation:
n1= 1.8 mol
p1= 27.3 atm
p2= 5.2 atm
n2= ?
n2= n1 * p2/p1
n2= 0.342 moles
Δn= n1-n2
Δn= 1.458 moles
A 0.7 kg lab cart moving to the right at 0.15 m/s collides with a 0.5 kg lab cart moving to the right at 0.10 m/s. After the collision, the 0.7 kg cart is moving to the right at 0.08 m/s. Calculate the velocity of the 0.5 kg lab cart after the collision.
Answer:
0.198 m/s
Explanation:
m1 = 0.7 kg, u1 = 0.15 m/s,
v1 = 0.08 m/s
m2 = 0.5 kg, u2 = 0.1 m/s
Let the speed of 0.5 kg is v2 after the collision.
By using the conservation of momentum
Momentum before collision = momentum after collision
m1 u1 + m2 u2 = m1 v1 + m2 v2
0.7 x 0.15 + 0.5 x 0.1
= 0.7 × 0.08 + 0.5 × v2
0.099 = 0.5 v2
v2 = 0.198 m/s
What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in air? Give your answer in units of tesla.
Answer:
The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]
Explanation:
Given that,
Distance = 0.36 m
Current = 3.8 A
We need to calculate the magnetic flux density
Using formula of magnetic field
[tex]B =\dfrac{\mu_{0}I}{2r}[/tex]
Where,
r = radius
I = current
Put the value into the formula
[tex]B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}[/tex]
[tex]B=2.11\times10^{-6}\ T[/tex]
Hence, The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]
A good baseball pitcher can throw a baseball toward home plate at 87 mi/h with a spin of 1710 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 ft path is a straight line.
Explanation:
First, convert 87 mi/h to ft/min.
87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min
The time to reach the home plate is:
t = 60 ft / 7656 ft/min
t = 0.00784 min
The number of revolutions made in that time is:
n = 1710 rev/min × 0.00784 min
n = 13.4 rev
Rounding to 2 significant figures, the ball makes 13 revolutions.
The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature inside the furnace?
Answer:
Temperature, T = 1542.10 K
Explanation:
It is given that,
The black body radiation emitted from a furnace peaks at a wavelength of, [tex]\lambda=1.9\times 10^{-6}\ m[/tex]
We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.
[tex]\lambda\propto \dfrac{1}{T}[/tex]
or
[tex]\lambda=\dfrac{b}{T}[/tex]
b = Wein's displacement constant
[tex]\lambda=\dfrac{2.93\times 10^{-3}}{T}[/tex]
[tex]T=\dfrac{2.93\times 10^{-3}}{\lambda}[/tex]
[tex]T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}[/tex]
T = 1542.10 K
So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.
Using Wien's Displacement Law, the temperature inside a furnace emitting radiation that peaks at a wavelength of 1.9 x 10^-6 m is approximately 1525 Kelvin.
Explanation:The temperature of the furnace can be determined using Wien's Displacement Law, which states that the product of the temperature of a black body and the wavelength at which the radiation it emits is most intense is approximately equal to 2.898 x 10^-3 m.K. In this specific case, the peak wavelength of the energy emitted from the furnace is given as 1.9 x 10^-6 m. Therefore, the temperature inside the furnace can be calculated using the formula: T = Amax / λ, where Amax = 2.898 x 10^-3 m.K and λ = 1.9 x 10^-6 m. Calculating from the given formula, T ≈ 1525 K. Therefore, the temperature inside the furnace is approximately 1525 Kelvin.
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A 50-kg crate is initially at rest on a horizontal floor. A person pushes horizontally with a constant force of 250 N. The maximum static frictional force between the floor and the crate is 275 N. What is the frictional force on the crate while the person pushes on it?
Answer:
250 N
Explanation:
m = Mass of cart = 50 kg
F = Pushing Force = 250 N
g = Acceleration due to gravity = 9.81 m/s²
μ = Coefficient of friction
[tex]F_s =\text{Static frictional force between the floor and the crate}= 275\ N[/tex]
N = Weight of cart = mg
N = 50×9.81
N = 490.5 N
[tex]F_s =\mu N\\\Rightarrow \mu=\frac{F_s}{N}\\\Rightarrow \mu=\frac{275}{490.5}\\\Rightarrow \mu=0.56[/tex]
Here it can be seen that the pushing force is less than the static frictional force so the crate will not move
∴ The frictional force on the crate while the person pushes on it is 250 N
The three-dimensional motion of a particle on the surface of a right circular cylinder is described by the relations r = 2 (m) θ = πt (rad) z = sin24θ (m) Compute the velocity and acceleration of the particle at t=5 s.
Answer:
[tex]V_{rex}=75.65m/s[/tex] and [tex]a_{res}=0[/tex] at t=5 secs
Explanation:
We have r =2m
[tex]\therefore \frac{dr}{dt}=0\\\\=>V_{r}=0[/tex]
Similarly
[tex]=>V_{\theta }=\omega r\\\\\omega =\frac{d\theta }{dt}=\frac{d(\pi t)}{dt}=\pi \\\\\therefore V_{\theta }=\pi r=2\pi[/tex]
Similarly
[tex]=>V_{z }=\frac{dz}{dt}\\\\V_{z}=\frac{dsin(24\pi t)}{dt}\\\\V_{z}=24\pi cos(24\pi t)[/tex]
Hence
at t =5s [tex]V_{\theta}=2\pi m/s[/tex]
[tex]V_{z}=24\pi cos(120\pi)[/tex]
[tex]V_{z}=24\pi m/s[/tex]
[tex]V_{res}=\sqrt{V_{\theta }^{2}+V_{z}^{2}}[/tex]
Applying values we get
[tex]V_{res}=75.65m/s[/tex]
Similarly
[tex]a_{\theta }=\frac{dV_{\theta }}{dt}=\frac{d(2\pi) }{dt}=0\\\\a_{z}=\frac{d^{2}(sin(24\pi t))}{dt^{2}}\\\\a_{z}=-24^{2}\pi^{2}sin(24\pi t)\\\\\therefore t=5\\a_{z}=0[/tex]