Answer:
Explanation:
they both involve nuclear reactions that produce energy, but the application are not the same
A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have to be increased for the pressure to be 1.5p_{1} ,
In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.
So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:
[tex]\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}[/tex]Then, let's calculate the proportion:
[tex]\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}[/tex]Now, converting back to Celsius, we have:
[tex]\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}[/tex]So the temperature would be 166.5 °C.
Block A in (Figure 1) has mass 0.900 kg , and block B has mass 3.00 kg . The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.35 m/s .
Part A: What is the final speed of block A?
Part B: How much potential energy was stored in the compressed spring?
(a) The final speed of block A is determined as 4.5 m/s.
(b) The potential energy that was stored in the compressed spring is 11.85 J.
What is the final speed of block A?
The final speed of block A is determined by applying the principle of conservation of linear momentum as follows;
Pa = Pb
where;
Pa is the momentum of block APb is the momentum of block Bmv (block A) = mv (block B)
(0.9 kg)(v) = (3 kg)(1.35 m/s)
0.9v = 4.05
v = 4.05/0.9
v = 4.5 m/s
The potential energy stored in the compressed spring is calculated as follows;
Apply the principle of conservation of energy.
U = K.E
where;
K.E is the kinetic energy of the blocksU = ¹/₂mv² (A) + ¹/₂mv² (B)
U = ¹/₂(0.9)(4.5²) + ¹/₂(3)(1.35²)
U = 11.85 J
Thus, the potential energy that was stored in the compressed spring is determined by applying the principle of conservation of energy.
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The Diagram shows the forces involved as a student slides a water bottle across the desk in front of them to their friend. Based on the image, in which direction is there friction? (ignore the selected answer it’s random)
Answer:
Left.
Step-by-step explanation:
The force of friction opposes the motion of an object, causing moving objects to lose energy and slow down. Therefore, the friction goes to the left.
Hey! I really need help with this question please :)
Answer: B
Explanation:
The formula for calculating the efficiency of a heat engine is expressed as
Efficiency = useful work done/Heat energy supplied x 100
From the information given,
Heat energy supplied = 500
useful work done = 50
Efficiency = 50/500 x 100
Efficiency = 10%
What is held in orbit by the gravitational pull of earth
The international space station.
The moon.
All TV satellites.
All weather satellites.
All GPS satellites.
More than 4000 other artificial satellites.
Thousands of pieces of "space junk"
Answer:
The Moon.
Explanation:
The earths gravity holds the moon in place.
Two ropes support a 15.0 kg load between them. One rope points NW at an angle of 15.0 degrees to the horizontal. Second rope points NE at an angle of 20.0 degrees to the vertical. Determine the magnitude of the tension force in each of the ropes.
The tension in the first rope is 38.05 N and the tension in the second rope is 138.1 N.
What is the weight of the load?
The weight of the load due to the force of gravity is calculated as follows;
W = mg
where;
m is mass of the loadg is acceleration due to gravityW = 15 kg x 9.8 m/s²
W = 147 N
Since the weight of the load is acting downwards, the tension in each rope is calculated as follows;
The tension in the first rope, T1 = W sinθ
where;
θ is the angle of inclination above the horizontalT1 = 147 sin(15)
T1 = 38.05 N
The tension in the first rope, T2 = W sinθ
where;
θ is the angle of inclination above the horizontal = 90 - 20 = 70⁰T2 = 147 x sin(70)
T2 = 138.1 N
Thus, the tension in each rope is determined by calculating the vertical component of force in each rope.
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A player hits a ball 45 degrees above the horizontal 1.3m above the ground. It clears a 3m wall 130m away. What is the minimum initial velocity the ball can clear the wall?
Explanation:
I had this question last year, let me check my book if i could find it.
A person is at the top of a tower. He takes a segment of a string which measures 30 cm long when at rest and hooks his 3 kg sword at the end of it. The spring extends to 35 cm long. He will use this spring to get to the ground. What is the spring constant of the spring, and how much of the spring (measured at equibilirum) does he need in order to have a net force of 0 upon himself when he touches the ground? Assume he hangs the spring from a hook located exactly 30 m above the ground. Be certain to draw a free body diagram of the forces on him the moment he hits the ground.
The given problem can be exemplified in the following diagram:
To determine the constant of the spring we can use Hook's law, which is the following:
[tex]F=k\Delta x[/tex]Where:
[tex]\begin{gathered} F=\text{ force on the string} \\ k=\text{ string constant} \\ \Delta x=\text{ difference in length} \end{gathered}[/tex]Now, we solve for "k" by dividing both sides by the difference in length:
[tex]\frac{F}{\Delta x}=k[/tex]The force on the string is equivalent to the weight attached to it. The weight is given by:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Substituting in the formula for the constant of the spring we get:
[tex]\frac{mg}{\Delta x}=k[/tex]Now, we substitute the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{35\operatorname{cm}-30\operatorname{cm}}=k[/tex]Before solving we need to convert the centimeters into meters. To do that we use the following conversion factor:
[tex]100\operatorname{cm}=1m[/tex]Therefore, we get:
[tex]\begin{gathered} 35\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.35m \\ \\ 30\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.30m \end{gathered}[/tex]Substituting in the formula we get:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.35m-0.30m}=k[/tex]Solving the operations:
[tex]588\frac{N}{m}=k[/tex]Therefore, the constant of the spring is 588 N/m.
what is the general importance of water?
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision.. A)each car has half the impulse of before B)each car has double the impulse of before C)each car have the same impulse D)each car has an impulse in ratio to its mass.
We are given the following information.
The collision is elastic.
Car B has twice the mass of car A.
Recall that in an elastic collision, the momentum and the kinetic energy are conserved.
Impulse is basically the change in momentum.
Since car B has 2 times the mass of car A, the momentum of car B will be 4 times the momentum of car A.
This means that the impulse of car B will be greater than the impulse of car A.
Option D says that each car has an impulse in ratio to its mass meaning that a car with a larger mass will have a larger impulse and vice versa.
Therefore, we can conclude that after the collision, each car has an impulse in ratio to its mass.
You can hear sound from another room through a door that is slightly open because...the sound refracts as it goes from one room to another.the sound wave reflects off the air in the doorway.the sound diffracts as it goes through the opening.the sound is polarized as it goes through the narrow opening.
when sound wave go from small passages like the edge of a wall or from the slit of open door the sound wave diffracts.
So the 3rd option is correct option.
Tall pacific coast redwood trees can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia come moving when it reaches the ground if it dropped from the top of a 100 m tree?
Given data:
Height of the tree;
[tex]h=100\text{ m}[/tex]Initial velocity;
[tex]u=0\text{ m/s}[/tex]The velocity of sequoia when it reaches the ground is given as,
[tex]v=\sqrt[]{u^2+2gh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(0\text{ m/s})^2+2\times(9.8\text{ m/s}^2)\times(100\text{ m})} \\ \approx44.27\text{ m/s} \end{gathered}[/tex]Therefore, sequoia will reach the ground with a velocity of 44.27 m/s.
An interesting question is “In what direction is the dot (representing a particle in the medium)moving at the instant shown above?” The correct answer is “_______”. The way to understandthat is to imagine the pulse an instant later. The wave will have moved a bit to the right.Therefore, since the particle is still on the wave, and can only move up or down, it must be______.
We will have the following:
The correct answer is vertically.
The way to understand that is to image the pulse an instant later. The wave will have moved a bit to the right. Therefore, since the particle is still on the wav, and can only move up or down, it must be lower.
Hafthor bjornson broke the deadlift record in April 2020, lifting 501kg. A) How much weight (in N) did he lift?B) How hard was the floor pushing up on the weights when they were on the floor?
Given, the mass that Hafthor Bjornson lifted, m=501 kg
A)
The weight is given by the product of the mass and the acceleration due to gravity.
Thus the weight lifted by him is,
[tex]\begin{gathered} W=mg \\ =501\times9.8 \\ =4909.8\text{ N} \end{gathered}[/tex]Thus the weight he lifted is 4909.8 N
B)
When the weight is on the floor the force applied by the floor on the weights is equal to the weight itself. This force is called the normal force.
Thus the force applied by the floor on the weights is 4909.8 N
a car goes from 32 m/s to a complete stop in 4.8 seconds. calculate the average stopping force of the car if has a mass of 2500 kg
The average stopping force is 16,500 N
Initial velocity of car (v₁)= 32m/s
Final velocity (v₂) = 0m/s
Time to stop= 4.8 seconds
Mass of car= 2500 kg
we need to apply the concept of laws of motion
Acceleration of car (a)= Change in velocity/time
a= v₂-v₁/t
a= 0-32/4.8
a= -6.6 m/s² ( deceleration)
Force= mass x acceleration
Force= 2500x 6.6
Force= 16500 N
Therefore the average stopping force is 16,500 N.
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What it mean for the brightness of bulbs in parallel if the potential difference across each one is the same as the potential difference across the battery?A. Not enough infoB. All the sameC. Decrease for each oneD. Increase for each one
B. All the same
Explanation
Total voltage of a parallel circuit has the same value as the voltage across each branch:
in the image, the voltage across R1 is the same as the voltage across R2,
Step 1
Increasing the voltage increases the brightness of the bulb. it means the brigthness depends on the voltage (also the brigthness depends on the current), so as the potential difference is the same, we can conclude the brigthness is the same, In a parallel circuit the voltage for each bulb is the same as the voltage in the circuit. Unscrewing one bulb has no effect on the other bulb.
so the answer is
B. All the same
I hope this helps you
What is the net force on an object with an applied force of 800N (right) and friction resisting at 750 N (left)?1 1550 N left2 1550 N right3 50 N left4 50 N right
Given,
The applied force, F=800 N
The friction, f=750 N
Friction is a force that opposes the motion of an object. Thus the net force will be equal to the difference between the applied force and the friction. As the applied force is greater than the frictional force, the net force will be in the same direction as the applied force, that is to the right.
Thus the net force is given by,
[tex]F_n=F-f[/tex]On substituting the known values,
[tex]\begin{gathered} F_n=800-750 \\ =50\text{ N} \end{gathered}[/tex]Therefore the net force on the object is 50 N to the right. Thus, the correct answer is option 4.
Which resistors in the circuit must always have the same current?A.B and CB.A and BC.C and DD.A and D
ANSWER
D. A and D
EXPLANATION
Two resistors have the same current if they are connected in series. As we can see in the schematic, resistors B and C are connected in parallel, so they don't have the same current - unless they have the same resistance.
Resistors A, D, and the equivalent of the parallel resistors (B and C) are connected in series, so they always have the same current.
Hence, of these options, resistors A and D always have the same current.
a ball starts from rest. It rolls down a ramp and reaches the ground after 8 seconds. It's final velocity when it reaches the ground is 14.0 meters/second. What is the initial velocity and acceleration?
A ball starts from rest such that initial velocity, u=0, and final velocity, v = 14 m/s
and the time duration, t = 8 seconds.
To find initial velocity and acceleration, a.
As the ball is at rest, thus initial velocity is zero.
Acceleration is given by the formula,
[tex]a=\frac{v-u}{t}[/tex]Substituting the values in the above equation, we get
[tex]\begin{gathered} a=\frac{14-0}{8} \\ =1.75m/s^2 \end{gathered}[/tex]Hence the acceleration is 1.75 m/s^2
Based on the circuit voltage and the wattage consumption,determine the approximate ampere rating of the followingappliances. Remember amps = watts divided by voltage.a = w÷ VRound to the nearest whole amp.1. AC Compressor on a 240 volt line and using 5,000 watts, amps =_____2. baseboard heater on a 120 volt line and using 1,200 watts, amps =_____3. vacuum cleaner on a 120 volt line and using 500 watts, amps =______4. blender on a 115 volt line and using 300 watts, amps5. toaster on a 120 volt line using 1,100 watts, amps =_____
Given:
1.
The voltage of AC compressor is V = 250 V
The power of the AC compressor is P = 5000 W
2.
The voltage of the baseboard heater is V = 120 V
The power of the baseboard heater is P = 1200 W
3.
The voltage of the vacuum cleaner is V = 120 V
The power of the vacuum cleaner is P = 500 W
4.
The voltage of the blender is V = 115 V
The power of the blender is P = 300 W
5.
The voltage of the toaster is 120 V
The power of the toaster is P = 1100 W
Required:
1. The approximate ampere rating of the AC compressor.
2. The approximate ampere rating of the baseboard heater.
3. The approximate ampere rating of the vacuum cleaner.
4. The approximate ampere rating of the blender.
5. The approximate ampere rating of the toaster.
Explanation:
1. The approximate ampere rating of the AC compressor can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{P}{V} \\ =\frac{5000}{240} \\ =20.833\text{ A} \\ \approx21\text{ A} \end{gathered}[/tex]2. The approximate ampere rating of the baseboard heater can be calculated as
[tex]\begin{gathered} I=\frac{1200}{120} \\ =\text{ 10 A} \end{gathered}[/tex]3. The approximate ampere rating of the vacuum cleaner can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{500}{120} \\ =4.2\text{ A} \\ \approx4\text{ A} \end{gathered}[/tex]4. The approximate ampere rating of the blender can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{300}{115} \\ =2.6\text{ A} \\ \approx3\text{ A} \end{gathered}[/tex]5. The approximate ampere rating of the toaster can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{1100}{120} \\ =9.2\text{ A} \\ \approx\text{ 9 A} \end{gathered}[/tex]Final Answer:
1. The approximate ampere rating of the AC compressor is 21 A.
2. The approximate ampere rating of the baseboard heater is 10 A.
3. The approximate ampere rating of the vacuum cleaner is 4 A.
4. The approximate ampere rating of the blender is 3 A.
5. The approximate ampere rating of the toaster is 9 A.
A 1.4N friction force slows a block to a stop after sliding 7m. How much work was done by the friction force
Answer:
9.8J
Explanation:
The work done by the friction force can be calculated as
W = Fd
Where F is the friction force, and d is the distance that the block slide.
So, replacing F = 1.4 N and d = 7 m
W = (1.4N)(7 m)
W = 9.8 J
Therefore, the work done by the friction was 9.8J against the movement of the block
a cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00s. wht is its acceleration?
The given value of the speed of cheetah is,
[tex]v=30ms^{-1}[/tex]The time during the speed v is,
[tex]t=7\text{ s}[/tex]The relation between the acceleration, speed and time is,
[tex]a=\frac{v}{t}[/tex]Substituting the known values,
[tex]\begin{gathered} a=\frac{30}{7} \\ a=4.3ms^{-2} \end{gathered}[/tex]Thus, the value of the acceleration is 4.3 meter per second squared.
Answer:
30×7=210m it is easyokok
What does a scientist mean when he or she says that an object is at rest
Answer:
I does not change position with respect to its surroundings with time
What is the frequency of a photon of EMR with a wavelength of 2.55x10*³m?1.18x1011 Hz8.50x10 12 Hz7.65x105 Hz1.18x105 Hz
In order to solve this equation, we will need to use the formula
[tex]f=\frac{c}{\lambda}[/tex]where f = frequency, c is the speed of light and lamda is wavelength
c = 3x10^8 m/s
lamda = 2.55x10^-3 m
f = (3x10^8)/(2.55x10^-3) = 1.18x10^11 1/s
A 244 kg motorcycle is travelling with aspeed of 14.7 m-s-1A) Calculate the kinetic energy (in J) of themotorcycle.B) If the speed of the motorcycle is increasedby a factor of 1.6, by what factor does itskinetic energy change?C) Calculate the speed (in m-s-1) of themotorcycle if its kinetic energy is 1/3 of thevaluefound in (a).
Given data:
* The mass of the motorcycle is m = 244 kg.
* The speed of the motorcycle is u = 14.7 m/s.
Solution:
(A). The kinetic energy of the motorcycle is,
[tex]K_1=\frac{1}{2}mu^2[/tex]Substituting the known values,
[tex]\begin{gathered} K_1=\frac{1}{2}\times244\times(14.7)^2_{} \\ K_1=26362.98\text{ J} \end{gathered}[/tex]Thus, the value of kinetic energy is 26362.98 J.
(B). If the speed of the motorcycle is increased by a factor of 1.6,
[tex]\begin{gathered} v=14.7\times1.6 \\ v=23.52\text{ m/s} \end{gathered}[/tex]Thus, the kinetic energy of the motorcycle becomes,
[tex]\begin{gathered} K_2=\frac{1}{2}mv^2 \\ K_2=\frac{1}{2}\times244\times(23.52)^2 \\ K_2=67489.23\text{ m/s} \end{gathered}[/tex]Dividing K_2 by K_1,
[tex]\begin{gathered} \frac{K_2}{K_1}=\frac{67489.23}{26362.98} \\ \frac{K_2}{K_1}=2.56 \end{gathered}[/tex]Thus, the kinetic energy is increased by the factor of 2.56.
(C). The 1/3 of the kinetic energy in the first part is,
[tex]\begin{gathered} K=\frac{1}{3}\times K_1 \\ K=\frac{1}{3}\times26362.98 \\ K=8787.66\text{ J} \end{gathered}[/tex]Thus, the speed of the motorcycle with the kinetic energy K is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2_{}_{} \\ 8787.66=\frac{1}{2}\times244\times v^2 \\ 8787.66=122\times v^2 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v^2=\frac{8787.66}{122} \\ v^2=72.03 \\ v\approx8.5\text{ m/s} \end{gathered}[/tex]Thus, the speed of the motorcycle is 8.5 m/s.
The figure shows a person whose weight is W = 607 N doing push-ups. Find the
normal force exerted by the floor on (a) each hand and (b) each foot, assuming that
the person holds this position.
The normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
Torque is defined as force times the distance from the line of force that is perpendicular to it.
(a) Each hand's typical response is, let's say, N1. For hands, the overall typical response is 2N1.
Balance the torque around the foot now.
torque is applied in a counterclockwise direction
Now ,
2N1 x 1.25 = W x 0.84.
2N1 × 1.25 = 607 × 0.84
N1 = (607 x 0.84)/(2x1.25)
N1 = 203.95N.
(a) The normal response for each foot is, let's say, N2, and the total normal response for feet is 2N2.
Now, distribute torque among the hands.
Anticlockwise torque is equal to clockwise torque.
Now,
2N2 x 1.25 = W x 0.41
2N2 × 1.25 = 607 × 0.41
N2 = (607× 0.41 )/2× 1.25
N2 = 99.55N
Hence,the normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
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When the buoyant force on an object is equal to or greater than its weight, the object __
When the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
What is buoyant force?
Buoyant force is the upward force exerted on an object that is fully or partly immersed in a fluid.
This upward force is also called Upthrust.
According to Archimedes' principle which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
An object will accelerate if its upthrust is greater than its weight, but will reach an upward terminal velocity when upthrust is equal to weight plus drag force.
Thus, when the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
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Two cars in opposite directions were going at 32 mph before a collision. They had a head on inelastic collision, i.e. the two cars stuck together afterward. The common speed of the combined piece right after the collision is 20 mph. The mass of Car 1 was 2,000 lb. Car 2 was heavier. The mass of Car 2 was ____ lb.
The mass of Car 2 was 3000 lb.
We need to apply the concept of conservation of momentum.
The velocity of both cars= 32mph
Combined velocity = 20mph
Mass of Car 1= 2000 lb
According to the conservation of momentum
M1V1+ M2V2= (M1+M2)
2000x32- (-M2x 32)=20(2000+M2)
64000+32M2=40000 +20M2
24000= 8M2
M2= 3000lb
Therefore the mass of Car 2 is 3000lb.
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Which of the following terms represents the number of waves passing a given point each second?Frequency⊝Wavelength⊝Amplitude⊝Velocity⊝CLEAR ALL
Frequency is defined as number of waves or vibration per unit time.
Therefore, option (a), frequency is the number of waves passing a given point each second, is the correct choice.
Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.
We are given that a jet is traveling with a speed of 78.6 m/s and travels a distance of 919m. We are asked to determine the constant acceleration when the jet stops. To do that we will use the following formula:
[tex]v^2_f=v^2_0+2ax[/tex]Where:
[tex]\begin{gathered} v_f=\text{ final speed} \\ v_0=\text{ initial speed} \\ a=\text{ acceleration} \\ x=\text{ distance traveled} \end{gathered}[/tex]Since the jet stops, this means that the final speed is zero. We will solve for the acceleration "a" in the formula. First, we will eliminate the term for the final speed since it is zero:
[tex]0=v^2_0+2ax[/tex]Now we will subtract the initial speed squared from both sides:
[tex]-v^2_0=2ax[/tex]Now we will divide by "2x" from both sides:
[tex]\frac{-v^2_0}{2x}=a[/tex]Now we replace the known values:
[tex]\frac{-(78.6\frac{m}{s})^2}{2(919m)}=a[/tex]Solving the operations:
[tex]-3.36\frac{m}{s^2}=a[/tex]Therefore, the magnitude of the acceleration is 3.36. Since the jet is deaccelerating in the direction due south, the direction of the acceleration is due north.