A student goes skateboarding a few times a week what is the dependent variable
The dependent variable in this scenario is the outcome or result that you are trying to measure or analyze based on the student's skateboarding activity.
Since the student goes skateboarding a few times a week, the dependent variable could be any aspect related to their skateboarding experience or its effects.
Examples of possible dependent variables could include:
1. Improvement in skateboarding skills (e.g., measured by tricks learned, levels of proficiency).
2. Physical fitness (e.g., measured by changes in endurance, strength, or flexibility).
3. Time spent skateboarding per session.
4. The number of skateboarding injuries or accidents.
5. Overall enjoyment or satisfaction with skateboarding.
6. Changes in stress levels or mood before and after skateboarding sessions.
7. Social interactions and friendships formed through skateboarding.
The specific dependent variable would depend on the research question or hypothesis you are investigating in relation to the student's skateboarding activity.
Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?
The speed of the white puck immediately after the collision is 2.6 m/s.
Two pucks are equal masses.
Speed of black puck = 1.5 m/s
According to given figure,
We need to calculate the speed of the white puck immediately after the collision
Using law of conservation of momentum
Put the value into the formula according to figure
Hence, The speed of the white puck immediately after the collision is 2.6 m/s.
A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.10 m/s. How much heat energy is dissipated to the environment?
We know that Total energy is conserved.
Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy
Initial kinetic energy = 0 ( magnet is at rest initially)
Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J
Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J
A poorly constructed room is suffering from a pipe leakage problem. The leaked pipes are continuously flooding the 90-m2 room such that the water level in the room increases at a steady rate of 1.2 cm/hr (a) How much water in L/min should be pumped out of the room to keep the water level cnstat (b) How much water in L/min should be pumped out of the room to reduce the water level by 4 cm/hr?
(a): should be pumped out of the room 18 L/min to keep the water level constant.
(b): should be pumped out of the room 78 L/min to reduce the water level by 4 cm/hr.
S= 90 m²
rate= 1.2 cm/hr = 0.012 m/hr = 0.0002 m/min
Water leak= S*rate= 90 m² * 0.0002 m/min
Water leak= 0.018 m³/min * 1000 L/m³
Water leak= 18 L/min (a) Water should be pumped out to keep the level constant.
By the rule of 3:
1.2 cm/hr ------------- 18 L/min
(4+1.2) cm/hr -------- x= 78 L/min (b) Water should be pumped out to reduce the level by 4 cm/hr.
In the calorimetry experiment which energy will be calculated during the heat exchange if water is used?
From the coffee cup to the thermometer
The assumption behind the science of calorimetry is that the energy gained or lost by the water is equal to the energy lost or gained by the object under study. So if an attempt is being made to determine the specific heat of fusion of ice using a coffee cup calorimeter, then the assumption is that the energy gained by the ice when melting is equal to the energy lost by the surrounding water. It is assumed that there is a heat exchange between the iceand the water in the cup and that no other objects are involved in the heat exchanged. This statement could be placed in equation form as
Qice = - Qsurroundings = -Qcalorimeter
The role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any error analysis of a calorimetry experiment must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings. And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings.
The energy calculated while dealing with the calorimeter experiment are the latent heat of vaporization, latent heat of fusion and the heat required to change the temperature of the substances.
The calorimeter works on the principle of conservation of energy. The amount of heat given by one part of the system is equal to the amount of heat gained by another part provided that the calorimeter does not loss any heat to the environment.
Consider that ice is mixed with water at some temperature. Then the water being at higher temperature losses heat to the ice at lower temperature. The ice gains the heat from the water and the system reaches an equilibrium at which the solution of ice and water has the same amount of energy at a particular temperature.
The different types of energies dealt with in the calorimetry experiment are as follows:
Latent heat of fusion:
The amount of energy required by a body when it is melted from its frozen state or freezes from its melted state is termed as the latent heat of fusion.
The small amount of ice is mixed with water in a calorimeter. Here, the ice requires the latent heat of fusion that leads to the melting of ice and converts it into water.
Latent heat of vaporization:
The amount of heat required to convert one gram of liquid to vapor without raising its temperature is known as latent heat of vaporization.
The water is boiling at in a calorimeter. Here, the water requires latent heat of vaporization which leads to the vaporization of water and convert it into vapors.
Thus, the latent heat of fusion, latent heat of vaporization and the heat required to change the temperature of the substance are the energies measured with the calorimeter.
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.
a) The total time of the trip is 4.05 h.
b) The average speed of the car is 12.35 mi/h.
c) The total time of the trip is 1.69 h.
a) The equation of traveled distance for a car traveling at constant speed is the following:
x= v · t
x = traveled distance.
v = velocity.
t = time.
Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":
x/v = t
So, the time it takes the car to travel the first half of the distance will be:
t1 = 25.0 mi / 7.00 mi/h
And for the second half of the distance:
t2= 25.0 mi / 52.00 mi / h
The total time will be:
total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h
total time = 4.05 h
The total time of the trip is 4.05 h.
b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:
a.s = d/t
a.s = 50 mi / 4.05 h
a.s = 12.35 mi/h
The average speed of the car is 12.35 mi/h
c) Let's write the equations of traveled distance for both halves of the trip:
For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:
7.00 mi/h = d1/t1
Solving for d1:
7.00 mi/h · t1 = d1
For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.
52.00 mi/h = d2/t2
52.00 mi/h · t2 = d2
We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time: