Which property describes the number of waves that move past a point each second.
A) Speed
B) Frequency
C) Period
D) Wavelength
Answer:B) Frequency
Explanation:
Answer:
B) Frequency
Explanation:
matter makes up all ? and ?
Which event is an example of condensation?
A. Wet clothes are drying on a clothesline.
B. A mirror fogs up when someone takes a hot shower.
C. Water drips from an icicle on the edge of a roof.
O D. Rain turns to sleet as it nears the ground.
Answer:
the answer to your question is b a mirror fogs up when someone takes a hot shower.
The process of conversion of gaseous water to its liquid form is called condensation. The fogs formed on the mirror from the hot shower is an example of condensation.
What is condensation?Condensation is the process of cooling up of water vapor to liquid water. Water vapor condenses when it cools. Condensation can be best understood from the reason behind raining.
When water vaporizers from resources and cools from the sky, and the vapor condenses to form liquid droplets. Similarly we can observe water droplets in the window pane due to the similar effect.
The water vapor arises from the hot shower condenses in the atmosphere and forms the drops on the mirror. Therefore, the mirror fogs up by the hot shower is an example of condensation.
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1. Which of the following provides a correct answer for a problem that can be solved using the kinematic equations?
A A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 32 m.
B A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 16 m/s
C A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's travels a
distance of 16 m
A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 24 m/s
E A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 30 m.
1 of 14
2
3
5 6 7 8
9
Answer:
12
Explanation:
12
What is the name of the kind of stretch that involves stretching as far as you can and then holding for 10-30 seconds
Question 2 options:
PNF
ballistic
dynamic
static
Answer:
Static stretching.
Explanation:
It is static stretching because it is a form of stretching which u can do actively for a period of time and you hold position for about 30 to 60 seconds which allow the muscles and connective tissues to lengthen. It is done after work out with out movement in order to loosen up muscles so as to gain flexibility.
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the decay constant and half-life T1/2; (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30 hours after it is prepared?
Answer:
(a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.
Explanation:
Given that,
Activity [tex]R_{0}=10\ mCi[/tex]
Time [tex]t_{1}=4\ hours[/tex]
Activity R= 8 mCi
(a). We need to calculate the decay constant
Using formula of activity
[tex]R=R_{0}e^{-\lambda t}[/tex]
[tex]\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})[/tex]
[tex]\lambda=0.0000154\ s^{-1}[/tex]
[tex]\lambda=1.55\times10^{-5}\ s^{-1}[/tex]
We need to calculate the half life
Using formula of half life
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}[/tex]
Put the value into the formula
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}[/tex]
[tex]T_{\dfrac{1}{2}}=44.719\times10^{3}\ s[/tex]
[tex]T_{\dfrac{1}{2}}=11.3\ hr[/tex]
(b). We need to calculate the value of N₀
Using formula of [tex]N_{0}[/tex]
[tex]N_{0}=\dfrac{3.70\times10^{6}}{\lambda}[/tex]
Put the value into the formula
[tex]N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}[/tex]
[tex]N_{0}=2.38\times10^{11}\ nuclei[/tex]
(c). We need to calculate the sample's activity
Using formula of activity
[tex]R=R_{0}e^{-\lambda\times t}[/tex]
Put the value intyo the formula
[tex]R=10e^{-(1.55\times10^{-5}\times30\times3600)}[/tex]
[tex]R=1.87\ mCi[/tex]
Hence, (a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.
Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs.
Which statement about work and power describes Hiro’s actions?
He did more work running than walking.
He did more work walking than running.
He had more power running than walking.
He had more power walking than running.
Answer:
C) he had more power running than walking
Explanation:
saw it on a quizlet. not 100% sure tho
Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.
What is power?In physics, power is the amount of energy transferred or converted per unit of time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.
Power is related to other quantities; for example, the power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle.
The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft
Since we know that Hiro will run on stairs to minimize the time of reaching his home so if time decreases then he has to increase his power as power is the ratio of work and time.
Hence Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.
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A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.
Normal
Gravity
Applied
Friction
Tension
Air resistance
Answer:
Normal, Gravity, Friction, and Air Resistance.
Explanation:
When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:
Normal force: It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.
Gravity: Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.
Friction: Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.
Air resistance: air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.
Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."
can you help me doing an essay about actual self
Answer:
yes absolutely
Explanation:
why not
Three toy boats with the same mass were in a lake. Two boats were moving and one was stopped. Each boat got bumped by another boat, but not in the same direction. All the boats changed speed as a result of being bumped. Use the information in the diagram to answer. Which toy boat exprienced the strongest force when it was bumped ? How do you know ?
The answer is " The orange and gray toy boats experienced the strongest force because both gained the same force as they sped up the blue boat lost force so it slow down
Explanation:
Trust me I did this one for an assignment
The boat that will experience the strongest force is the boat that has the highest speed.
According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.
F = ma
[tex]F = \frac{mv}{t}[/tex]
where;
F is the force experienced by the objectv is the velocity of the objectt is the time of motionThe force experienced by the boats is directly proportional to the velocity of their motion. Since the boats have the same mass, the force experienced by each boat will depend on the speed with which the boat moves.
Thus, the boat that will experience the strongest force is the boat that has the highest speed.
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The planet's hawks and block their near each other in the door again system the dworkin's have very advanced technology and a dork and scientist want to increase the pull of gravity between the 2 planets which proposals with the scientists make to accomplish this goal check all that aply
Answer: A,B, and E
Explanation: Just checked I got them right:)
2
How is acceleration related to force when mass is constant, according to Newton's second law of motion?
A. The acceleration is directly proportional to the net force,
OB
The acceleration is inversely proportional to the net force.
C. The acceleration is inversely proportional to the square root of the net force.
OD
The acceleration is directly proportional to the square root of the net force.
Reset
Next Question
Answer:
A
Explanation:
The acceleration of an object is directly proportional to its net force.
[tex]a = \frac{f}{m} [/tex]
Answer:
Correct choice: A. The acceleration is directly proportional to the net force
Explanation:
Newton's Second Law of Motion
According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force F and inversely proportional to the object's mass m:
[tex]\displaystyle a=\frac{F}{m}[/tex]
Correct choice: A. The acceleration is directly proportional to the net force
A force of 500 N acts for a time interval of 0.001 second on an object of mass 0.20 kg that was initially at rest. Calculate the final velocity of the object after the force acts
Answer:
100
Explanation:
The final velocity of the object after the force acts is 2.5 meters per second.
To calculate the final velocity of the object after the force acts, we can use Newton's second law of motion, which states that:
Force = Mass × Acceleration
We need to find the acceleration of the object first. Since the object was initially at rest, its initial velocity (u) is 0 m/s.
The force acting on the object is 500 N, and the mass of the object is 0.20 kg. We can rearrange the equation to find the acceleration (a):
Acceleration (a) = Force / Mass
a = 500 N / 0.20 kg
a = 2500 m/s²
Now, we can use the kinematic equation to find the final velocity (v) of the object:
Final velocity (v) = Initial velocity (u) + (Acceleration × Time)
The initial velocity (u) is 0 m/s (since the object was initially at rest), and the time interval (t) is 0.001 second.
Final velocity (v) = 0 m/s + (2500 m/s² × 0.001 s)
v = 0 m/s + 2.5 m/s
v = 2.5 m/s
Hence, the final velocity of the object after the force acts is 2.5 meters per second.
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Why should you never do an exercise such as the bench press without a spotter?
Answer:
you could hurt yourself
Explanation:
Answer
You should never do an exercise such as the bench press without a spotter because you could hurt yourself or push your self too much.
Explanation:
16. True/False: Protons are exchanged between objects by induction
protons are present in sodium atom.
a 11
b 10
C 12
d 9
11
Explanation:
There are 11 protons in a sodium atom
In economics, _________ is the amount of a resource that firms and producers are willing and able to provide to the marketplace
sectors are the amount of resources
Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force
Answer:
Resultant is 152 N at 28.5 degrees south to the 100 N force
Explanation:
Two ropes are attached a wagon one horizontal to the west with a tension force of 75N and other east and at an angle of 38 degrees upward from horizontal with a lension force of 125 N Find the components of the net force on the cart Show your work
Answer:
The components of the net force acting on the cart are;
Horizontal force, Fₓ ≈ 23.5 N acting east
The vertical force, [tex]F_y[/tex] ≈ 76.96 N
Explanation:
The parameters given in the question are;
The magnitude of the force on the rope acting horizontally to the west = 75 N
The magnitude of the force on the other rope east acting at an angle to the horizontal = 125 N
The angle to the horizontal of the force on the other rope east to the horizontal = 38°
Therefore, we have;
The horizontal component of the force acting east = 125 × cos(38°) ≈ 98.5 N
The net sum of the horizontal forces, ∑Fₓ noting that the horizontal forces are acting in opposite direction is given as follows;
∑Fₓ = 75 - 125 × cos(38°) = 75 - 98.3 = -23.5N
Given that the difference in forces is negative, and hat we took the force acting east as negative, we have that the net horizontal force is acting east with a horizontal component of 23.5N
The net vertical force, ∑F[tex]_y[/tex], is given as follows;
∑F[tex]_y[/tex] = 125 × sin(38°) ≈ 76.96 N
Therefore, the components of the net force acting on the cart are;
Horizontal force, Fₓ ≈ 23.5 N, the vertical force, [tex]F_y[/tex] ≈ 76.96 N
The components of the net force on the cart are 23.5 Newton and 76.96 Newton.
Given the following data:
Tension force (horizontal) due west = 75 NTension force (horizontal) due east = 125 NAngle of inclination = 38°To find the components of the net force on the cart:
First of all, we would determine the horizontal component of the force acting due east.
[tex]F_y = Fcos(\theta)\\\\F_y = 125cos(38)\\\\F_y = 125 \times 0.7880\\\\F_y = 98.5 \; Newton[/tex]
Next, we would find the resultant horizontal forces:
[tex]\sum F_y = F + F_y\\\\\sum F_y = 75 + (- 98.5)\\\\\sum F_y = -23.5 \;Newton[/tex]
Therefore, the horizontal component of the net force acts in the opposite direction due east.
Horizontal component of the net force = 23.5 Newton
For the vertical component of the net force:
[tex]\sum F_x = Fsin(\theta)\\\\\sum F_x = 125cos(38)\\\\ \sum F_x = 125 \times 0.6157\\\\ \sum F_x = 76.96\; Newton[/tex]
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A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has
a mass of 0.44 kg, what is the acceleration of the soccer ball?
A. 27.3 m/s2
B. 21.3 m/s2
C. 110 m/s2
D. 104 m/s2
Answer:
C. 110 m/s2
Explanation:
Force = Mass x Acceleration
Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:
Force/Mass = (Mass x Acceleration)/Mass
Acceleration = Force/Mass
Now we just have to plug in our values and calculate!
Acceleration = 48.4/0.44
Acceleration = 110m/s/s
It is option C. 110 m/s2
Hope this helped!
help do number 16 and 17
Answer:
16 is d and i have no idea for 17
4. Which of the following statements is correct?
A Mass and weight are different names for the same thing
B The mass of an object is different if the object is taken to the Moon
C The weight of a cer is one of the forces acting on the car.
D The weight of a chocolate beris measured in kilograms
Answer:
Explanation:
A: wrong. Mass and weight are different.
B: Wrong. The mass here and the mass on the moon are the same. The weight, which is Mass * the acceleration is equal to weight.
C: Correct.
D: Wrong. Weight is not measured in Kg. Mass is.
A student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons. What is the force of the wall pushing back on the student? What is the acceleration of the student as she moves away from the wall?
Answer:
F=-100N; a=1.3m/s^2
Explanation:
Force is being made by student, so wall counteracts that force by not moving so it is equally opposite.
The force of the wall pushing back on the student would be 100 Newtons, and the acceleration of the student as she moves away from the wall would be 1.33 m/s²
What is Newton's third law of motion?Newton's third law of motion state that for every action force there exists a complementary reaction force that balances it.
As given in the problem a student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons,
The force of the wall pushing back the student = 100 Newtons
The acceleration of the student = 100/75
=1.33 m/s²
Thus, The student would be pushed back against the wall by a force of 100 Newtons, and she would accelerate away from the wall at a rate of 1.33 m/s².
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Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4 kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based on the data given below. Total energy is the sum of gravitational potential energy and kinetic energy. In this problem, round gravity to: g = 10 m/s^2.
Answer:
its should be 2.0 and 4.5 on it
chemical formulawhat is the chemical formula
Answer: A chemical formula is a way of showing and representing information about a chemical proportion of atoms that constitute a particular chemical compound or molecule
Determine the weight of a 10.0-kg person who is running with a speed of 5.0 m/s. Enter a
numerical answer.
Answer:
weight=mass*gravity
weight=10.0*10
so weight=100N
1. According to Newton's third law of motion, how are action
and reaction forces related?
2. How is momentum conserved?
3. Suppose you and a friend, who has exactly twice your mass,
are on skates. You push away from your friend. How does the
force with which you push your friend compare to the force
with which your friend
pushes you? How do your
accelerations compare?
4. Thinking Critical be ly comparing and Contrasting
Which has more momentum, a 250-kg dolphin swimming
at 6 m/s, or a 450-kg manatee swimming at 2 m/s?
(need 1-4 answered ASAP)
Answer:
you couldn't do this on your own or search it up on google
Give reasons.
a. A stretched catapult has potential energy.
b A stone on ground has zero energy.
C. Rolling a drum is work against friction.
d. Lifting a stone is work against gravity.
e. Work and energy both have same unit.
Yodelin has fifty quarters and dimes. Their total value is $9.80. Which of these systems of equations can be used to find the number of quarters (q) and
dimes (d) Yodelin has?
Answer:
q + d = 50
25q + 10d = 980
Explanation:
The equation that can be employed to determine the number of quarters(q), as well as, dimes(d) that Yodelin has would be:
q + d = 50...(1)
25q + 10d = 980...(2)
Multiplying (1) by 25
25q + 25d = 1250 ...(3)
25q + 10d = 980...(4)
subtracting (4) from (3)
15d = 270
d = 18
q = 50 - 18
= 32
The system of the equations could be
q + d = 50
25q + 10d = 980
Calculation of the equation and the number of quarters and dimes:Since Yodelin has fifty quarters and dimes. Their total value is $9.80.
So here the equations be
q + d = 50...(1)
25q + 10d = 980...(2)
Now we have to Multiplying (1) by 25
So,
25q + 25d = 1250 ...(3)
25q + 10d = 980...(4)
Now
subtracting (4) from (3)
15d = 270
d = 18
So,
q = 50 - 18
= 32
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7. A car is moving at 50.0 mph when the driver applies brakes. Determine the distance it
covers before coming to a halt. Coefficient of static friction between the tires and surface of
the road is 0.514. Mass of the car is 1000 kg.
Answer:
The distance to come to a halt is approximately 49.53 meters
Explanation:
Thee given parameters are;
The speed of the car, v = 50 mph = 22.35 m/s
The mass of the car, m = 1000 kg
The coefficient of friction, = 0.514
The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N
The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J
The work done by the brake = Force of the brake × Distance, d, to come to halt
By conservation of energy, we have;
The work done by the brake = The initial kinetic energy of the car
∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt
The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt
∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters
The distance to come to a halt ≈ 49.53 meters.
The car will cover 49.65 m distance before stopping due to application of brake.
Given data:
The mass of car is, M = 1000 kg.
The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.
The coefficient of static friction is, [tex]\mu = 0.514[/tex].
The given problem can be solved using the third kinematic equation of motion. Which is,
[tex]v^{2}=u^{2}+2as[/tex] ....................................................(1)
Here, v is the final speed, v = 0 (Because car is finally stopping)
s is the distance covered before stopping and a is the magnitude of acceleration.
Now, since frictional force opposes the motion. Then,
[tex]F = f\\\\ma = \mu \times m \times g\\\\a = 0.514 \times 9.8\\\\a =5.03 \;\rm m/s^{2}[/tex]
Substituting the values in equation (1) as,
[tex]0^{2}=23.35^{2}+2(-5.03)s\\\\s = \dfrac{2500}{2 \times 5.03}\\\\ s=49.65 \;\rm m[/tex]
Thus, the car will cover 49.65 m distance before stopping due to application of brake.
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