Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?
Answer:
I think it would be 50 I am not really sure
Explanation:
I think you would have to divid 1000 by 20 Again I'm not sure
A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead wanted the car to achieve twice as much speed at the bottom of the ramp, how high should the ramp be compared to the first case
Answer:
It must be 4 times high.
Explanation:
Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.This means, that at any time, the following must be true:ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)⇒ [tex]m*g*h = \frac{1}{2} * m*v^{2}[/tex]
Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
[tex]m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}[/tex] (1)
Now, let's do v₂ = 2* v₁Replacing in (1) we get:[tex]m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}[/tex] (2)
Dividing (2) by (1), and rearranging terms, we get:h₂ = 4* h₁A car moving at a speed of 8.0 m/s merges onto the highway and accelerates at a rate of
3.0 m/s2. How fast is it moving after travelling for 54 meters?
[tex]\large\underline{\underline{\red{\sf \blue{\longmapsto} Step\:-\:by\:-\:step\: Explanation:-}}}[/tex]
As per the data in Question,
Initial velocity of car = 8m/s (u)Final velocity of car = ? (v) .Acceleration of the car = 3.0m/s².Distance travelled by the car = 54m .Using the third equation of motion :-
[tex]\boxed{\red{\bf\purple{\dag}\:\:2as\:=\:v^2\:-\:u^2}}[/tex]
Substituting the respective values ,
⇒ 2as = v² - u² .
⇒ 2 × 3 × 54 = v² - 8²
⇒ 324 + 64 = v² .
⇒ v² = 388.
⇒ v = √388 .
⇒ v = 19.69 m/s ≈ 20m/s .
Hence the required answer is 19.69 m/s.
Which statement best describes the direction of force shown by the magnetic field lines around a bar magnet?
A. Away from both the magnet's north and south pole
B. From the magnet's north to its south pole
C. From the magnet's south pole to its north pole
D. Toward both the magnet's north and south pole
Answer: I'm pretty sure the answer is B
Explanation: I'll check after the test
:Okay yes the answer is 100% B
The direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet. Option B is correct.
Magnetic Field Lines:
These are the imaginary lines that are used to represent the strength of the magnetic field.
The magnetic field lines starts from the North pole and ends up in the South pole.The strength of the magnetic field is directly proportional to the number of lines per unit area.The strength of the magnetic field is the highest near the north pole.Therefore, the direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet.
To know more about Magnetic Field Lines,
https://brainly.com/question/17011493
My parrot has a mass of 1.33kg, what is it's weight here on earth
Answer:
Your parrot, from earth, that weighs 1.33kg is 1.33kg on earth. as far as i'm aware there is only one earth and everything always weighs the same on one planet as it did on that same planet.
Explanation:
why do you think we need to study physics
Answer:
The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics.
You pull a wagon with a force of 20 N. The wagon has a mass of 10 kg. What is the wagon's acceleration?
Answer:
The answer is 2 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question
f = 20 N
m = 10 kg
We have
[tex]a = \frac{20}{10} \\ [/tex]
We have the final answer as
2 m/s²Hope this helps you
Answer:.
Explanation:.
A doctor refers a patient for an MRI image in the frontal plane. Into which
divisions would this plane divide the image body part?
A right and left
B proximal and distal
C anterior and posterior
D superior and inferior
Answer:
C anterior and posterior
Explanation:
c
Can someone please help me with this following question, If you could visit Pangaea what animals would you find
A). Penguins
B). Mammals
C). Dinosaurs
D). Eagles
Answer:
C). Dinosaurs
Explanation:
Experiment 1: A study is done to determine which of two fuel mixtures allows a rocket to travel farther over a period of time. Rocket A, which requires additional equipment to keep it stable, is used to test one fuel mixture, and rocket B is used to test the other. Both rockets are identical aside from their mass. The results indicate that rocket B traveled farther than rocket A over the same period of time. Experiment 2: A double-blind experiment is performed to test whether a new drug is effective in lowering blood pressure. A random sample of subjects with high blood pressure is assigned to two groups. One group receives the new drug and the other group does not. Neither group is permitted to take any other medications during the experiment or to change their lifestyles in any way. The results of the experiment show that the drug is effective in lowering blood pressure.
Identify the experiment in which confounding occurs and the reason for its occurrence.
a. Neither experiment has a confounding variable.
b. Experiment 1 has a confounding variable related to the fuel mixtures. Varying the fuel mixture could skew the results of the study and should be kept constant.
c. Experiment 2 has a confounding variable related to the type of experiment. A double-blind experiment may increase the risk of the placebo effect and possibly skew the results.
d. Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.
e. Experiment 2 has a confounding variable related to the subjects used. Choosing a sample of subjects with high blood pressure instead of individuals with different blood pressure levels may confuse the results.
Answer: D
Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.
This is the answer to the question because:
Both experiments do have a confounding variable.Experiment 1 doesn't have to stay constant.A double-blind experiment will not do anything to the placebo.High blood pressure people will not make the results confusing.The answer has to be the option D. Hope this helps you!
An object is dropped from a 45 m high building. At the same time, another object is thrown
upward with a velocity of 8.5 ms 1. How high above the ground will the two objects meet?
(With work please)
Answer:
-92.33 (meaning the objects will not meet above the ground).
Explanation:
We can use the kinematic equation displacement = initial velocity*time + 1/2*acceleration*time^2.
We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:
x = 0*t + 1/2*(-9.8)*t^2+45
x = 8.5*t + 1/2*(-9.8)*t^2
We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:
0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2
-4.9*t^2 +45 = 8.5*t + -4.9*t^2
45 = 8.5*t
t = 45/8.5 ≈5.294
Now, we can plug t as 5.294 into any of the equations above to solve for x:
x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33
That means, the objects will not meet above the ground.
(iii) Why do right angle mirrors produce three images of the object?
Explanation:
The two mirrors inclined to each other formed the first two images with are of the same size as the object while the third mirror is produced from the intersection of rays that emanated during the production of the first two images to produce a third image which is smaller than the object and there making the total number of images to be 3.
Hence this mirrors produces 3 images due to the third image formed from the intersection of the rays that produces the first two images.
The formula that relates the image produced by inclined mirror and the angle of inclination is expressed as:
number of images n = 360/θ - 1
θ is the angle of inclination of the two mirrors
n is the number of images
If the mirrors are inclined at right angles, then θ = 90°
Substitute into the formula;
n = 360/90 -1
n = 36/9 -1
n = 4-1
n = 3
A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
Why is it better to use the metric system, rather than the English system, in scientific measurement?
A. The English system uses one unit for each category of measurement.
B. The metric system uses one unit for each category of measurement.
C. The English system uses consistent fractions that are multiples of 10.
D. The metric system utilizes a variety of number conversions.
A. The English system uses one unit for each category of measurement.
Answer:
A
Explanation:
A marathon is 26 mi and 385 yd long. Estimate how many strides would be required to run a marathon. Assume a reasonable value for the average number of feet/stride.
Answer:
According to the University of Iowa, the average length of a stride is 5ft.
Now, the total distance of the marathon is:
26mi and 385yd.
Let's transform that distance into ft.
1mi = 5280ft
Then:
26mi = 26*5280ft = 137,280ft
1yd = 3ft
then:
385yd = 385*3ft = 1,155ft.
Then the total distance of the marathon, in ft, is:
D = 137,280ft + 1,155ft = 138,435 ft.
Now the average number of strides needed will be equal to the quotient between the total distance of the marathon and the distance traveled in each stride.
N = 138,435ft/5ft = 27,687.
1. A speed boat is racing across a lake at 25 meters per
second when its motor burns out. It then slowly
comes to a stop over the next 45 seconds. What was its
acceleration?
v = u + a t
where u = initial velocity (25 m/s), v = final velocity (0), a = acceleration, and t = time (45 s). So
0 = 25 m/s + a (45 s)
a = (-25 m/s) / (45 s)
a ≈ -0.56 m/s²
Please help. :( I am horrendously bad at Physics...
1) Why is the mass of an NFL player important to the game?
- Essay question.
2) What causes objects to accelerate?
A. Mass
B. Force (my answer)
C. Inertia
D. Velocity
3) Impulse is related to the amount of force acting on a football for a period of time. What will happen to a football if the impulse it experiences is increased?
- Essay question
4) There is a gravitational force between the Earth and the Moon. Which mass pulls harder?
A. The Earth pulls harder
B. The moon pulls harder
C. This is a trick question, they pull with the same force. (My answer)
5) The moon goes around the Earth while the Earth seems to be not affected. Why is the Earth not really moved by the force of gravity applied by the moon? (Hint: Consider the mass of the Earth and inertia)
- Essay question
6) Imagine a force is applied to a toy to make it roll across the floor. How would the acceleration of the toy car change if the same force is applied for a longer period of time?
A. Acceleration would be smaller
B. Acceleration would be greater
C. Acceleration would be the same (my answer. ?)
D. Acceleration would change randomly
7) Which gravitational force is greater? The Earth's gravitational force on you or your gravitational force on Earth?
A. The Earth's gravitational force on me is greater
B. My gravitational force on the Earth is greater
C. The two forces are equal
I know this is a bit long, but I really appreciate anyone who could help. <3
Answer:
1) I believe The mass of a football player is important because. If you are a football player and you are not able to tackle ,push , throw the ball very far- exedra, then you will not be able to play because football includes many factors of weigh. therefor you have to have a greater mass or you will not be able to play football or you wont be very good which can lead to being kicked off the team. So in order to be able to play football you must be of a certain topic of weight.
Explanation:
.
A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30 m/s. What is the force exerted by the water jet
Answer:
Explanation:
mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water
A = 3.14 x .005²
= 785 x 10⁻⁷ m²
mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg
momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .
Rate of change of momentum = 706.5 x 10⁻⁴
Let force be F
F - mg = 706.5 x 10⁻⁴
F = mg + 706.5 x 10⁻⁴
F = 23.55 x 10⁻⁴ x 9.8 + 706.5 x 10⁻⁴
= 937.3 x 10⁻⁴ N .
5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]E(r)_1 = [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]
Now when the outer diameter is doubled, the radius(b) is also doubled
So
[tex]E(r)_2 = [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]
Now
[tex]\frac{E(r)_2}{E(r)_1} = \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} = \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]
Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
Two boxes of masses 3M and 5M are attached by a massless rope. They are being pulled to the right with a constant force of P = 800 N, which allows them to just overcome static friction, with a μs= 0.70 between the floor and the boxes.
a. Find M.
b. Find the Tension in the rope between the two boxes.
Answer:
a) about 14.577 kg
b) 300 N
Explanation:
b) In order for the acceleration to be the same for each mass, the 800 N force must be divided between the boxes in proportion to their mass. That is, the net force on the 5M mass must be 5/8 of the total force, or 500 N. Then the tension in the rope is 800 N -500 N = 300 N, which is 3/8 of 800 N.
Tension: 300 N
__
a) The total mass is 8M, and the total normal force on the floor is ...
F = ma = (8M)(9.8 m/s^2)
The friction force is 0.7 times this, and is equal to the 800 N force pulling on the boxes.
800 N = (8M)(9.8 m/s^2)(0.7)
M = 800/(8·9.8·0.7) kg ≈ 14.577 kg
Determine the Northward and the Eastward components of a 75 m long displacement vector which points in a direction of 27^o E of N.
Answer:
Eastward = 34.05m
Northward = 66.8m
Explanation:
First, let's define our coordinate axis.
I will choose the North as the positive y-axis and the East as the positive x-axis.
Now we will work with polar coordinates, r and θ and remember that if we want to recover the rectangular coordinates, we need to compute:
x = r*cos(θ)
y = r*sin(θ)
(remember that θ is measured counterclockwise from the positive x-axis)
Now we have:
"... 75 m long displacement vector..."
r = 75m
"...points in a direction of 27° E of N..."
This means that the measure is from the positive y-axis, clockwise (so this is not equivalent to θ).
Now, remember that the angle between the positive y-axis and the positive x-axis is 90°.
Then we will have that θ is the complementary angle to 27°, or:
θ = 90° - 27° = 63°
Now we have θ and r, then we can calculate the rectangular components:
x = 75m*cos(63°) = 34.05m (this is the Eastward component)
y = 75m*sin(63°) = 66.8m (this is the Northward component)
What is the moment of inertia I of an object that rotates at 13.0 rev/min13.0 rev/min about an axis and has a rotational kinetic energy of 16.0 J?
Answer:
The moment of inertia of the object is 17.276 kilogram-square meters.
Explanation:
According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ([tex]K_{R}[/tex]), measured in joules, is represented by the following formula:
[tex]K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}[/tex] (Eq. 1)
Where:
[tex]I_{G}[/tex] - Moment of inertia with respect to center of mass, measured in kilogram-square meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now we clear the moment of inertia:
[tex]I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}[/tex]
If we know that [tex]K_{R} = 16\,J[/tex] and [tex]\omega \approx 1.361\,\frac{rad}{s}[/tex], then the moment of inertia of the object is:
[tex]I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{G} =17.276\,kg\cdot m^{2}[/tex]
The moment of inertia of the object is 17.276 kilogram-square meters.
The moment of inertia of the object will be "17.276 kg/m²".
Moment of inertiaRotational Kinetic energy, [tex]K_R[/tex] = 16 J
Angular speed, ω = 1.361 rad/s
By using the Rotation Physics, the relation will be:
→ [tex]K_R[/tex] = [tex]\frac{1}{2}[/tex] × [tex]I_G[/tex] × ω²
the,
The moment of inertia be:
→ [tex]I_G[/tex] = [tex]\frac{2\times K_R}{\omega^2}[/tex]
By substituting the values, we get
= [tex]\frac{2\times 16}{(1.361)^2}[/tex]
= [tex]\frac{32}{(1.361)^2}[/tex]
= 17.276 kg.m²
Thus the above answer is correct.
Find out more information about Kinetic energy here:
https://brainly.com/question/25803184
Which two types of energy does a book have as it falls to the floor
Answer:
kinetic and potential energy
Explanation:
Which material could be represented by the image above? (SC.8.P.8.1)
A. Diamonds
B. Milk
C. Oxygen
D. Wood
Answer:
a. diamonds
Explanation:
solid is the matter that has definite shape and definite volume. One good example is a diamond because contains particles that are strongly attract to each other.
A jet plane lands with a speed of 100 m/s and can
accelerate at a maximum rate of 5.00 m/s2 as it comes to
rest. (a) From the instant the plane touches the runway,
what is the minimum time needed before it can come to
rest? (b) Can this plane land on a small tropical island
airport where the runway is 0.800 km long?
Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
[tex]v_{f}=v_{i}-(a*t)[/tex]
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
[tex]x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}[/tex]
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land
What is Newton's 2nd law in a simple definition
Answer:
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
Explanation:
Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?
after
Variables:
Equation and Solve:
Answer:
We are given:
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
Solving for initial velocity:
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]