Answer:
Explanation:
The efficiency of the heat engine is less than 100%. The heat engine converts the thermal energy from the burning of fuel to the mechanival energy. Thus, some of the energy gets lost as waste heat.
Hence, this waste heat is used up to heat up the interior of the car and person feels warmer in the car.
Thus, the waste heat generated in the engine is used for the heating of the car interior.
Final answer:
Running the heater in a car on a cold winter day doesn't cost much additional energy beyond running the fan because the heat generated by the engine is mostly released into the environment instead of being used for work in the car.
Explanation:
Heat engines convert thermal energy into work. In a car engine, heat is produced when fuel is burned, and this heat is converted into kinetic energy to move the car and power its systems. Running the heater in a car on a cold winter day doesn't cost much additional energy beyond running the fan because the heat generated by the engine is mostly released into the environment instead of being used for work in the car.
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on:________.a) the mass of the sphere.
b) the radius of the sphere.
c) both the mass and the radius of the sphere.
d) neither the mass nor the radius of the sphere.
The angular velocity of a solid uniform sphere rolling without slipping down an inclined plane does not depend on the mass or the radius of the sphere; it depends on the height from which it rolls down and gravitational acceleration only.
The question asks what determines the angular velocity of a solid uniform sphere as it rolls without slipping down an inclined plane. It specifically queries whether this angular velocity depends on the mass (M) of the sphere, the radius (R) of the sphere, both, or neither.
The answer is that the angular velocity of a rolling sphere at the bottom of the incline does not depend on the mass of the sphere or the radius of the sphere. According to the principles of conservation of energy and the dynamics of rolling motion, all objects regardless of their mass or radius, will roll down an incline and reach the bottom with the same angular velocity if they start from rest, provided that they do not slip and there is no air resistance. This is because the potential energy lost is completely converted into kinetic energy (both translational and rotational), and the mechanical energy of the system remains constant.
The angular velocity of the sphere at the bottom of the incline depends on d) neither the mass nor the radius of the sphere.
When the sphere rolls without slipping, there is a relationship between its linear velocity (v) and angular velocity (ω) given by the equation v = ωR.
Total mechanical energy of the sphere = translational kinetic energy + rotational kinetic energy + potential energy.
At the bottom of the incline, all potential energy will have converted into kinetic energy:
Initial Potential Energy (U) = mghTranslational Kinetic Energy (Kt) = (1/2)mv²Rotational Kinetic Energy (Kr) = (1/2)Iω², where I for a solid sphere is (2/5)mR²By energy conservation:
mgh = (1/2)mv² + (1/2)Iω²Substituting I and the relation v = ωR gives us:
mgh = (1/2)mv² + (1/2)*(2/5)mR²ω²Solving for ω:
mgh = (1/2)mv² + (1/5)mR²ω²Since v = ωR, ω² = v²/R²mgh = (1/2)mv² + (1/5)m(v²)7gh = v²v = √(7gh)ω = v/R = √(7gh)/RFrom the above steps, it's evident that the angular velocity ω depends on the height h the sphere rolls down, and the acceleration due to gravity g, but not on the mass M or radius R of the sphere.
Therefore, the correct answer is: d) neither the mass nor the radius of the sphere.
If you are caught outdoors in a thunderstorm, why shouldn’t you stand under a tree? Can you think of a rea- son why you should not stand with your legs far apart? Or why lying down can be dangerous? (Hint: Consider the electric potential difference.)
Answer:
Because its dangerous.
Explanation:
During lighting strikes, there is discharge or energy transfer(electrons) from the clouds to the earth, these electrons flow through the path with least resistance between the cloud and the earth. Also the electric field around the tip of the leaves are strong, which makes trees a great target.
we as humans have lower resistance than trees, that is to say, the lighting may leave the tree and flow trough the body to the earth.
The tree and the ground around it are then raised to a high potential relative to the ground some distance away.
If you stand with your legs far apart, one leg on a higher-potential part of the ground than the other, or if you lie down with a potential difference between your head and your feet, you may find yourself a conducting path.
If it is also raining, the electricity may transfer down the wet tree to the wet ground and shock anyone standing near the tree.
During a thunderstorm, it is advised not to stand under a tree to avoid being struck by lightning, not to stand with legs far apart to reduce the electric potential difference, and not to lie down due to increased risk. Staying inside a car provides safety as it acts as a Faraday cage.
Explanation:Standing under a tree during an electrical storm is dangerous since lightning tends to strike the tallest object in an area, which could be the tree you're under, potentially causing serious injury or death. As for not standing with legs apart, this is because, in the event of a ground strike, electricity can travel through the ground. If your legs are far apart, there could be a significant electric potential difference between them, which can result in a stronger current passing through your body, leading to severe harm. Lying down increases your contact with the ground, and consequently, the risk of current flowing through your body from a ground strike is greater.
During thunderstorms, your car acts as a Faraday cage, which shields you from electric fields if a lightning strike occurs nearby. It's safest to remain inside the car with windows closed. However, it's critical to refrain from touching metal parts inside the car as lightning can transfer its charge through the car's metal frame.
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a uniform electric fiedl of magnitude E.
Answer:
Range, [tex]R = MV²/2QE[/tex]
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
This traction loss occurs in the rear wheels of a vehicle. a) Braking-induced traction loss b) Acceleration-induced traction loss c) Front wheel traction loss (skid) d) Rear wheel traction loss (skid) e) Driver-induced skids
Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.
Explanation:The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.
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Final answer:
Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.
Explanation:
The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.
With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.
Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.
2-lbm of water at 500 psia intially fill the 1.5-ft3 left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300°F. Determine the final pressure of water, in psia, and the total internal energy, in Btu, at the final state.
Explanation:
Formula for final volume of chamber if the partition is ruptured will be as follows.
[tex]V_{2}[/tex] = 1.5 + 1.5
= 3.0 [tex]ft^{3}[/tex]
As mass remains constant then the specific volume at this state will be as follows.
[tex]\nu_{2} = \frac{V_{2}}{m}[/tex]
= [tex]\frac{3.0}{2}[/tex]
= 1.5 [tex]ft^{3}/lbm[/tex]
Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.
[tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]
[tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]
[tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]
Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.
[tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]
1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]
[tex]x_{2}[/tex] = 0.229
Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F
and, [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia
Formula to calculate internal energy at the final state is as follows.
[tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]
= [tex]2(269.51 + 0.229 \times 830.45)[/tex]
= 920.56 Btu
Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.
A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between the two charges?
Answer:
Ep= 3.8 10⁵ N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ =+7.5 nC = +7.5*10⁻⁹C
q₂ = -2.0 nC = -2.0*10⁻⁹C
d₁ =d₂ = 1.5cm = 1.5 *10⁻²m = 0.015 m
Calculation of the electric fieldsat the midpoint (P) between the two charges
Look at the attached graphic:
E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .
E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₁ = k*q₁/d₁² = 9*10⁹ *7.5 *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep= E₁ + E₂
Ep= 3*10⁵ N/C + 0.8*10⁵ N/C
Ep= 3.8 10⁵ N/C
The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.
Explanation:The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.
Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.
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A solar eclipse that occurs when the new moon is too far from Earth to completely cover the Sun can be either a partial solar eclipse or a(n)_____________.
Answer:
ANULAR ECLIPSE
Explanation:
ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.
Sir Lance a Lost new draw bridge was designed poorly and stops at an angle of 20o below the horizontal. Sir Lost and his steed stop when their combined center of mass is 1.0 m from the end of the bridge. The bridge is 8.0 m long and has a mass of 2000 kg; the lift cable is attached to the bridge 5.0 m from the castle end and to a point 12 m above the bridge. Sir Lost’s mass combined with his armor and steed is 1000 kg.
Determine
(a) the tension in the cable and
(b) the horizontal and vertical force components acting on the bridge at the castle end.
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.
In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:
A)The Tension T is 42120N
B)The Horizontal force component is 18322.2N
C)The Vertical force component is - 4729N
Before starting the calculations we have to remember some concepts such as:
Rx = x component of Reaction forceRy = y component of Reaction force.T = TensionW = weight of bridgew = weight of Sir Lance a Lost and his steedτ = torqueFirst we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:
[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]
Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:
[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]
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In 1977, Kitty O’Neil drove a hydrogen peroxide–powered rocket dragster for a record time interval (3.22 s) and final speed (663 km/h) on a 402-m-long Mojave Desert track. Determine her average acceleration during the race and the acceleration while stopping (it took about 20 s to stop). What assumptions did you make?
Answer:
57.19461 m/s²
-9.20833 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]
The acceleration during the race is 57.19461 m/s²
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]
The acceleration while stopping is -9.20833 m/s²
The assumptions made here are:
The initial velocity of the dragster at the start of the race is zero
The acceleration is constant in both cases of acceleration.
The final velocity when it stops is zero
Motion is in a straight line path
A force of 14 N acts on a 5 kg object for 3 seconds.
a. What is the object’s change in momentum?
b. What is the object’s change in velocity?
Answer: a) 42Nm b) 8.4m/s
Explanation:
Impulse is defined as object change in momentum.
Since Force = mass × acceleration
F = ma
Acceleration is the rate of change in velocity.
F = m(v-u)/t
Cross multiply
Ft = m(v-u)
Since impulse = Ft
and Ft = m(v-u)... (1)
The object change in velocity (v-u) = Ft/m from eqn 1
Going to the question;
a) Impulse = Force (F) × time(t)
Given force = 14N and time = 3seconds
Impulse = 14×3
Impulse = 42Nm
b) The object change in velocity (v-u) = Ft/m where mass = 5kg
v-u = 14×3/5
Change in velocity = 42/5 = 8.4m/s
A bowl contains 7 red balls and 10 blue balls. A woman selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue?
Answer:
1050 possible selections
Explanation:
Number of red balls = 7
Number of blue balls = 10
Let blue balls be B and red balls be R
four balls are to be selected. At least three must be blue.
the first combination is 4blue 0red
second combination is 3blue 1red
(10C4*7C0 + 10C3*7C1)
210 *1 +120 *7
210+840
=1050 possibilities
There are 71 different selections possible if at least 3 balls must be blue.
Explanation:To find the number of different selections possible, we can consider the different scenarios where at least 3 balls are blue:
Scenario 1: Exactly 3 balls are blue and 1 ball is red. In this case, there are 10 ways to select the 3 blue balls (from the 10 blue balls available) and 7 ways to select the red ball (from the 7 red balls available). So, the total number of selections for this scenario is 10 * 7 = 70.Scenario 2: All 4 balls are blue. In this case, there is only 1 way to select all 4 blue balls. So, the total number of selections for this scenario is 1.Summing up the number of selections from both scenarios, we get 70 + 1 = 71 different selections possible.
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Where did Earth’s water come from?, select the two competing scientific theories supported by evidence that explain the formation of the hydrosphere.
Answer:
Explanation:
There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.
The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.
Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.
Explanation:There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:
Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.
Suppose a 58-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 655 N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
Answer:
You have to apply force of 78.43 N to pull the sled or person across the snow at constant speed
Explanation:
Given data
Sled force F₁=58 N
Person weight F₂=655 N
Coefficient of kinetic friction u=0.11
To find
Friction Force
Solution
To pull the sled across the snow means the force you apply must be equal to and opposite the force of frictional.
First we need to find the Total normal force
So
[tex]F_{Normal-Force}=F_{1}+F_{2}\\F_{Normal-Force}=58N+655N\\F_{Normal-Force}=713N[/tex]
Now the force of friction
[tex]F_{Friction-Force}=u_{coefficient}*F_{Normal-Force}\\F_{Friction-Force}=0.11*713N\\F_{Friction-Force}=78.43N[/tex]
So you have to apply force of 78.43 to pull the sled or person across the snow at constant speed
Force = mass x acceleration : __________.a. for any force, there is an equal and opposite reaction force. b. the orbit of each planet about the Sun is an ellipse with the Sun at one focus. c. a planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times. d. an object moves at constant velocity if there is no net force acting upon it. e. more distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2 =a3.
Answer:
a. for any force, there is an equal and opposite reaction force.
Explanation:
Newton third law of motion states that " for every action, there is an equal and opposite reaction.
Action force= reaction force =
M₁ * a₁ = M₂ * a₂
where M₁ = mass of object 1
a₁ = acceleration due to object 1
M₂ = mass of object 2
a₂ = corresponding acceleration due to object 2
this happen in everyday life of an individual. an example is seen in a groups of three man exerting a push force on a static car, the push force on the car make the car to exhibit some motion equivalent to the force applied by the three man.
Most active galactic nuclei are at large distances from us; relatively few nearby galaxies harbor active galactic nuclei. What does this imply?
Answer:
The milky way does not have an active galactic nuclei and active galactic nuclei reduce their activity as time pass.
Explanation:
An active galactic nucleus (AGN) is a region that is compact and at the galaxy's center. It has a extra high luminosity in which some observations indicate that the luminosity is not as a result of the presence of stars.
7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
Explanation:
Given
mass of children [tex]m_1=20\ kg[/tex]
[tex]m_2=30\ kg[/tex]
distance between two children [tex]L=3\ m[/tex]
suppose small child is at a distance of x m from pivot point
so torque of small child and heavier child must be equal
[tex]20\times (x)=30\times (3-x)[/tex]
[tex]2x=9-3x[/tex]
[tex]5x=9[/tex]
[tex]x=1.8\ m[/tex]
You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick annoyingly to you. If two of the spheres carry equal charges and repel with a force of 22 { mN} when they're 16 { mm} apart, what's the magnitude of the charge on each?
Answer:
The magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex].
Explanation:
Given that,
Force of repulsion between the charges, F = 22 mN
The distance between spheres, r = 16 mm = 0.016 m
It is mentioned that both the spheres carry equal charges. The force between charges is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}[/tex]
[tex]q=2.50\times 10^{-8}\ C[/tex]
So, the magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
Answer:
b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.
Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
The change in entropy of the gas is 374.39 J/K.
The given parameters:
Mass of the ice, m = 0.3071 kgTemperature of the gas, T = 273.15 KLatent heat of fusion of water, L = 333,000 J/kgThe heat of fusion of the ice is calculated as follows;
[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]
The change in entropy of the gas is calculated as follows;
[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]
Thus, the change in entropy of the gas is 374.39 J/K.
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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.20 m/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 43.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 5.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? Where is Henrietta when she catches the bagels?
Answer:
[tex]u_x=8.5454\ m.s^{-1}[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Explanation:
Given:
speed of walking of Henrietta, [tex]v_w=3.2\ m.s^{-1}[/tex]height of projection of projectile, [tex]h=43.9\ m[/tex]Horizontal distance between the window and Henrietta when the projectile was launched:[tex]r=3.2\times 5[/tex]
[tex]r=16\ m[/tex]
Since the projectile was thrown at the time when she had passed below the window 5 seconds ago.Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.
Now the time taken for the object to reach to Henrietta ignoring the height where she catches:
[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]
[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]
[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.
Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:
[tex]\Delta d=v_w\times t[/tex]
[tex]\Delta d=9.5782\ m[/tex]
Now the total horizontal distance from the window to Henrietta when the projectile reached her:
[tex]d=\Delta d+r[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Now the initial horizontal velocity of launch of the projectile:
[tex]u_x=\frac{d}{t}[/tex]
[tex]u_x=\frac{25.5782}{2.9932}[/tex]
[tex]u_x=8.5454\ m.s^{-1}[/tex]
A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2. at 10 s after launch the first stage of the rocket (now without fuel) is released. the second stage now had an acceleration of 6 m/s2
a) how high is the rocket when the first stage seperates?
b)how fast is the rocket moving upon first stage seperation?
c) what will be the maximum height attained by the first stage after seperation?
d) what will be the distance between the first and second stages 2 s after separation
Answer:
a) 200m
b) 40 m/s
c) 81.55m
d) 31.62m
Explanation:
Solution
a)
y = y0 + u×t+⅟2×a×t2 =
y0 = 0
u = 0
y = unknown
a = 4m/s2
t = time = 10 seconds
y = 0.5×4×102 = 200m
b) v = u + at
v = 0 + 4×10 = 40 m/s
c) v2 = u2 - 2×g×y
at maximum height v = 0
we have
402 = 2×9.81×y
Y =81.55m
d)
for the stage 2 we haace
y = y0 + u×t+⅟2×a×t2 =
y = 0 + 4×2+0.5×6×22 = 92m
for the stage one we have
y = 0+40×2-0.5×9.81×4= 60.38m
distance between the first and second stage 2s aftee separation = 92-60.38 = 31.62m
To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by:_________
(a) the worker’s applied force,
(b) the gravitational force on the crate, and
(c) the normal force exerted by the incline on the crate?
(d) What is the total work done on the crate?
Answer:
a. [tex]W_w=313.5\ J[/tex]
b. [tex]W_g=-155.312\ J[/tex]
c. [tex]F_N=222.045\ N[/tex]
d. [tex]W_t=313.5\ J[/tex]
Explanation:
Given:
angle of inclination of the surface,[tex]\theta=25^{\circ}[/tex]mass of the crate, [tex]m=25\ kg[/tex]Force applied along the surface, [tex]F=209\ N[/tex]distance the crate slides after the application of force, [tex]s=1.5\ m[/tex]a.
Work done by the worker who applied the force:
[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.
[tex]W_w=209\times 1.5[/tex]
[tex]W_w=313.5\ J[/tex]
b.
Work done by the gravitational force:
[tex]W_g=m.g\times h[/tex]
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
[tex]h=s\times sin\ \theta[/tex]
[tex]h=1.5\times sin\ 25^{\circ}[/tex]
[tex]h=0.634\ m[/tex]
So, the work done by the gravity:
[tex]W_g=25\times 9.8\times (-0.634)[/tex] ∵direction of force and displacement are opposite.
[tex]W_g=-155.312\ J[/tex]
c.
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \theta[/tex]
[tex]F_N=25\times 9.8\times cos\ 25[/tex]
[tex]F_N=222.045\ N[/tex]
d.
Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]
The time in hours it takes a satellite to complete an orbit around the earth varies directly as the radius of the orbit (from the center of the earth) and inversely as the orbital velocity. If a satellite completes an orbit 660 miles above the earth in 13 hours at a velocity of 32,000 mph, how long would it take a satellite to complete an orbit if it is at 1500 miles above the earth at a velocity of 38,000 mph?
To find the new orbital period of a satellite at 1500 miles above Earth with a velocity of 38,000 mph, we use a proportional relationship comparing it to a known orbit. We set up the ratio using the direct variation with the orbit's radius and inverse variation with the orbital velocity, then solve for the unknown orbital period.
Explanation:The duration of a satellite's orbit around Earth can be described as varying directly with the orbit's radius and inversely with the orbital velocity. To calculate the orbital period for a new satellite trajectory, we need to compare it with a known situation using a proportional relationship. Given that a satellite orbits 660 miles (or 1062 kilometers) above Earth in 13 hours at 32,000 mph, we can set up a ratio to determine the orbital period at 1500 miles (or 2414 kilometers) above Earth with a velocity of 38,000 mph.
The ratio for the original orbit is T1 / (r1 / v1) = T2 / (r2 / v2), where T is time, r is radius (distance from Earth's center to the satellite), and v is velocity. Using the provided values, we get:
13 / ((3963 + 660) / 32000) = T2 / ((3963 + 1500) / 38000)
Solving for T2 gives us the new orbital period. Considering that we should convert the altitude to the same units and use Earth's radius in miles (3963 miles), the final calculation will provide us with the new orbital period in hours.
Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is increasing when the radius is 4.70 in. Assume that π = 3.14.
Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
Which of these statements best describe a leader? A. A leader is any person who is a part of the management team. B. A leader is any person who is always the first to perform any business activity. C. A leader is a person who inspires other team members to reach the organizational goals D. A leader is any person who puts others’ interest before his or her own. E. A leader is the one who owns the business and runs it on a day to day basis.
Answer:
I think the answer is c
Explanation:
But it is kinda of opinionated
Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals
What are the qualities of leader?A leader is someone who motivates others to achieve the organization's objectives.
A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.
A leader is a person who inspires other team members to reach the organizational goals
Hence,statements C best describe a leader.
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A spectral line that appears at a wavelength of 321 nm in the laboratory appears at a wavelength of 328 nm in the spectrum of a distant object. We say that the object's spectrum is:
Answer:
We say that the object's spectrum is shifted.
Explanation:
Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
That shift can be used to find the velocity of the object by means of the Doppler velocity.
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
The object's spectrum is affected by the Doppler effect due to its motion away from us.
The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.
The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.
This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.
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A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass of fuel ejected is small compared to the mass of the rocket. For this distance, which of the following indicates the changes, if any, in the kinetic energy of the rocket, the gravitational potential energy of the Earth-rocket system, and the mechanical energy of the Earth-rocket system? System Gravitational System Potential Energy Rocket Kinetic Energy Mechanical Energy
(A) Increasing IncreasingIncreasing
(B) Increasing Increasing Constant IncreasingDecreasing Decreasing Decreasing Increasing Constant
Answer:
A) Increasing, Increasing, Increasing
Explanation:
The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.
Gravitational potential energy,
as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.
Therefore, mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.
Hence Option A is correct.
For the first kilometer of the rocket's ascent, the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system increase.
Explanation:The changes in the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system for the first kilometer of the rocket's ascent can be determined.
Since the mass of fuel ejected is small compared to the mass of the rocket, the rocket's kinetic energy will increase as it accelerates away from Earth. The gravitational potential energy of the Earth-rocket system will also increase as the rocket moves higher against the pull of gravity. Therefore, the changes in the three quantities are as follows:
Kinetic Energy: IncreasingGravitational Potential Energy: IncreasingMechanical Energy: Increasing_______ describes the relationship between a stimulus and its resulting sensation by proposing that the JND is a constant fraction of the stimulus intensity.
Answer:
Weber's Law
Explanation:
Weber’s law is an important law in psychology.
This law quantifies the perception of change within a provided stimulus.
This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.
Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.
Final answer:
Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.
Explanation:
The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.
To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.
This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.
50 J of work was performed in 20 seconds. How much power was used to perform this task? A. 0.4 W B. 2.5 W C. 4 W D. 24.5 W
Answer:
B. 2.5 W
Explanation:
Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).
Mathematically,
Power = Energy or work/Time
P = W/t ........................Equation 1
Where P = power used to perform the task, W = work, t = time.
Given: W = 50 J, t = 20 s.
Substitute into equation 1
P = 50/20
P = 2.5 W.
Hence the power is 2.5 W.
The right option is B. 2.5 W
The power used to perform the work of 50J in 20 seconds is 2.5 watts. This much power should be generated by it. The correct answer is B.
Given that 50 J of work was performed in 20 seconds,
The power used to perform a task, you can use the formula:
Power = Work / Time
Power = 50 J / 20 s
Power = 2.5 W
Therefore, the power used to perform the task is 2.5 W.
Therefore, the correct answer is B) 2.5 W.
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30 seconds of exposure to 115 dB sound can damage your hearing, but a much quieter 94 dB may begin to cause damage after 1 hour of continuous exposure. You are going to an outdoor concert, and you'll be standing near a speaker that emits 50 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
39.8 m ≈ 40 m
Explanation:
power (P) = 50 W
sound intensity level ([tex]p[/tex]) = 94 dB
the distance (r) can be gotten from the equation I = [tex]\frac{power}{4nr^{2} }[/tex] (take not that π is shown as [tex]n[/tex])
making r the subject of the formula we have r = [tex]\sqrt{\frac{power}{4nI} }[/tex] (take not that π is shown as [tex]n[/tex])
But to apply this equation we need to get the value of the intensity (I)
we can get the intensity (I) from the formula sound intensity level ([tex]p[/tex]) = 10 log₁₀[tex](\frac{I}{I'})[/tex] rearranging the above formula we have intensity (I) = [tex]I' x 10^{\frac{p}{10} }[/tex]I' = reference intensity = 1 x[tex]10^{-12} W/m^{2}[/tex]now substituting all required values into the formula for intensity (I) I = [tex]1 x 10^{-12} x 10^{\frac{94}{10} }[/tex] = 0.00251 [tex]W/m^{2}[/tex]now that we have the value of the intensity (I) we can substitute it into the formula for the distance (r)
distance (r) = [tex]\sqrt{\frac{power}{4nI} }[/tex]
r = [tex]\sqrt{\frac{50}{4x3.142x0.00251} }[/tex] = 39.8 m ≈ 40 m