Answer:
y = 6
Explanation:
Given parameters:
Given equation:
y = 2x + 10
And supposing; x = -2
Unknown:
y = ?
Solution:
We are going to solve this problem by substitution;
y = 2x + 10
since x = -2; input this into the equation to solve for y;
y = 2(-2) + 10 = -4 + 10 = 6
The value of y = 6
Will was riding his bike when a dog ran out in front of him. He slammed on his brakes. During this quick stop, some of the mechanical energy (his motion) was changed into
A) heat energy.
B) light energy.
C) kinetic energy.
D) gravitational energy.
Answer:
A
Explanation:
big brain
During this quick stop, some of the mechanical energy (his motion) was changed into heat energy.
What is energy?Energy is the ability to do work. There are different types of energy such as Heat energy, light energy, kinetic energy, and gravitational energy.
Heat is the energy that moves from one body to another when temperatures are different. Heat passes from the hotter to the colder body when two bodies with differing temperatures are brought together.
The joule is a unit of energy that serves as the SI unit for heat (J). The calorie (cal), which is defined as "the amount of heat necessary to raise the temperature of one gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius," is another common unit of heat measurement.
Therefore, During this quick stop, some of the mechanical energy (his motion) was changed into heat energy.
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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!
A moving object with a decreasing velocity covers distance during
each new second than it covered in the previous second.
A.the same
B.more
C.less
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 shortest path between two points is:
1) displacement
2) breadth
Answer:
the shortest path between two points is displacement..
Explanation:
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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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matter makes up all ? and ?
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 student fires a cannonball vertically upwards. The cannonball returns to the
ground after a 4.60s flight. Determine all unknowns and answer the following
questions. Neglect drag and the initial height and horizontal motion of the
cannonball. Use regular metric units (ie. meters).
How long did the cannonball rise?
unit
What was the cannonball's initial speed?
unit
What was the cannonball's maximum height?
unit
V
Answer:
(a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
Explanation:
Given that,
Time = 4.60 s
We need to calculate the initial velocity
Using equation of motion
[tex]v=u-gt[/tex]
Put the value into the formula
[tex]0=u-9.8\times4.60[/tex]
[tex]u=45.0\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=45\times4.60[/tex]
[tex]d=207\ m[/tex]
We need to calculate the maximum height
Using equation of motion
[tex]v^2=u^2-2gh[/tex]
Put the value into the formula
[tex]0=(45.0)^2-2\times9.8\times h[/tex]
[tex]h=\dfrac{(45.0)^2}{2\times9.8}[/tex]
[tex]h=103.3\ m[/tex]
Hence, (a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
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
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
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
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2. What can the staff member estimate for the original speed of the race car if it came to a stop during the skid?
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = [tex]\sqrt{734.22}[/tex]
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
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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can we add 2 atoms together? 3? How do particles combine to form the variety of matter one observes?
A water skier is towing by motorboat at a constant velocity of magnitude 15 km /h. The boat speed up, and after a short interval the skier is towed at a new constant velocity of magnitude 20 km/h. What is the net force on the skier when she is moving at 15 km/h? And at 20 km/h?
Answer:
The skier is experimenting a net force of 0 newtons in both cases. ([tex]v_{1} = 15\,\frac{km}{h}[/tex], [tex]v_{2} = 20\,\frac{km}{h}[/tex])
Explanation:
According to Newton's First Law, an object is in equilibrium when it is either at rest or moving at constant velocity, which means that net force is equal to 0 newtons.
Therefore, the skier is experimenting a net force of 0 newtons in both cases.
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:
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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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
protons are present in sodium atom.
a 11
b 10
C 12
d 9
11
Explanation:
There are 11 protons in a sodium atom
I would appreciate the help
Two classmates, Aisha and Brandon, want to attend two school activities
over the coming weekend. They have one parking pass between them. The
probabilities that the classmates will attend each event are shown in the
table.
Alsha
Brandon
0.65
0.94
Probability of attending
the Saturday activity
Probability of attending
the Sunday activity
0.80
0.43
They decide to let the person more likely to attend both events have the
parking pass. Assuming that attendance at one activity is independent of
attendance at the other, who is more likely to attend both activities?
A. Brandon. He has a 0.40 probability of attending both activities
O B. Brandon. He has a 0.69 probability of attending both activities
C. Aisha. She has a 0.52 probability of attending both activities.
Answer:
The answer is D.
Explanation:
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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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.
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
which of the following is the same as 1 second
Answer:
theres no
Explanation:
picture or anything so... ur question doesnt make sense
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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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."
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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A rocket takes off from Earth. It travels 825km in 75 seconds. What is the
speed of the rocket?
0 1 km/s
11 km/s
O 1.1 km/s
O 111 km/s
Answer:
11 km/s
Explanation:
v=s/t
v=825km/75s
v=11km/s