Answer:
The answer is 198.23 kg
Explanation:
To find the mass of an object given it's mass and acceleration we use the formula
where
m is the mass
F is the force
a is the acceleration
From the question
a = 92.54 m/s²
F = 18,344 N
We have
We have the final answer as
198.23 kg
Answer:
198.22
Explanation:
using m=f/a you divide the force by the acceleration to get the mass (18344 / 92.54) = 198.22
A car is pushed and travels 12 m. The same car is then pushed and it travels 15m. What
could have happened? Select 2 answers.
The mass was decreased
The force was increased
The force was decreased
The mass was increased
=========================================
Explanation:
If the force is kept the same, but the mass was decreased, then the car will travel further. The less mass an object has, the less inertia it has. Inertia is the tendency for the object to stay at rest if it's initially at rest (or the tendency for the object to stay in motion if already in motion).
If the mass is kept the same, but the force was increased, then the object will travel further than before. The same idea applies as discussed in the last paragraph.
------
If the force was decreased and mass kept the same, then there is less push, which results in the object traveling a shorter distance. If the mass is increased but the force kept the same, then the object moves a shorter distance because there's more inertia to overcome. There are more particles needed to be moved and they resist motion if at rest.
When are all the forces acting upon an object balanced?
Answer:
When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.
Explanation:
A 70 kg student is riding on a skateboard at 4m/s. A 60 kg student is also riding a
skateboard at 4 m/s. Which statement below demonstrates an understanding of
Newton's 1st Law of Motion?
A. Since both students are traveling at the same speed they must have the same
inertia
B. Since both students are traveling at the same speed they must have the same
mass
C. One student has more mass and therefore more inertia
A storm is moving east toward your house at an average speed of 35 km an hour if the storm is currently 80 km from your house and how much time do you expect it to arrive
Answer: 2 hours 17 minutes
Explanation: 80/35h = 16/7 = 2hrs 17 minutes
Waves do NOT carry.
A)Weight
B)Matter
C)Energy
D)Color
(Choose one)
The jogger ran 3km east.
4. Look at the bottom figure. Suppose that a car is 20 km due north of New York City. The car
travels north toward Albany until it is 100 km due north of New York City.
Answer: d
Explanation:
Which of the following tests uses calipers to assess body composition? A. skinfold test B. sit-and-reach C. arm hang D. curl-up Please select the best answer from the choices provided.
Answer:A.skinfold test
Explanation:
i just took the test
Answer:
A. Skinfold test
Explanation:
Edg 2020
A cows mass is 401 kg and a trucks mass is 832 kg. What is the difference between their weights . Answer please!!!
Answer:
403 kg is the diffrence
Determine the acceleration that an unbalanced force of 28 N gives to a 6.8 kg mass.
Answer:
4.12 m/s^2
Explanation:
unbalanced force means the presence of resultant force (net force)
[tex]net \: force \: = mass \times accleration[/tex]
f = 28N
m = 6.8kg
a = ?
? = 28/6.8
= 4.1176...
approximately = 4.12 m/s^2
PLEASE HELP 50 POINTS!!!!!!!!!!!!!!!
Andrew and Deshawn are teammates on a high school soccer team. They’ve been playing soccer together for years. They both play the center forward position. Some years, Andrew is better than Deshawn and serves as the team’s starter. Other years, Deshawn is the stronger player and gets the most playing time. They are both very competitive with each other. Although they know each other well, they have never been friends.
In practice yesterday, the team was running drills in the rain. Deshawn slipped in the mud and collided with Andrew. Andrew hurt his knee badly in the fall and will have to sit out for several weeks.
The next day, Andrew told everyone at school that Deshawn hurt him on purpose to get more playing time. Deshawn retaliated by sharing private information about Andrew on social media. Andrew’s best friend Mateo saw the collision. He knows it was a complete accident.
Pick the role of Andrew, Deshawn, or Mateo.
What are the main issues/problems from this person’s perspective?
What ethical issues are involved for this person?
Why are these ethical issues relevant for this person?
How should the person you chose handle the situation?
Answer:
1. I pick the role of Mateo.
2. The main problems from Mateo's perspective are: He knows that the collision was a complete accident and that Deshawn did not injure Andrew on purpose. Because Andrew told everyone that Deshawn injured him on purpose, Deshawn did something unethical by sharing private information about Andrew on social media.
3. The ethical issues are: Andrew took his hurt and frustration of being injured and missing play time out on Deshawn, so he slandered him by saying that Deshawn hurt him on purpose. Deshawn then retaliated by sharing personal information about Andrew on social media, which is also unethical.
4. These issues are relevant to Mateo because he is Andrew's best friend AND he saw the incident happen and knows it was an accident.
5. If I was Mateo, I would talk to Andrew and explain to him that I know he is mad that he is injured and can't play for several weeks, but telling everyone that Deshawn did it on purpose wasn't right or good sportsmanship. I would encourage Andrew to meet with Deshawn with me and have Andrew apologize to Deshawn for saying it was his fault, and then encourage Andrew to go and tell everyone the real story. I would also encourage Deshawn to apologize for posting private information about Andrew, have Andrew ask him to take it down, and then ask Deshawn to post another post saying why he did it and that he was sorry. I would also encourage them to work together, since they are both star players.
Explanation:
Two cars are traveling on a desert road between three consecutive poles, as shown in the
figure. After 5.1 s, they are side by side at the
next telephone pole. The distance between
the poles is 72.7 m.
1-Find the displacement of Car A after 5.1 s.(answer in units of m.)
2-Find the average velocity of Car A during 5.1
s. (answer in units of m/s.)
Answer:
SUGGGA
Explanation:
Consider the following debate between two students about their answer to the above question.
Student 1: I thought that whenever one object exerts a force on a second object, the second object also exerts a force that is equal in strength, but in the other direction. So even though Earth is bigger and more massive than the Moon, they still pull on each other with a gravitational force of the same strength, just in different directions.
Student 2: I disagree. I said that Earth exerts the stronger force because it is way bigger than the Moon. Because its mass is bigger, the gravitational force Earth exerts has to be bigger too. I think you are confusing Newton's third law with the law of gravity. Do you agree or disagree with either or both of the students?
Answer:
I agree with student 1
Explanation:
This is because, the magnitude of the gravitational force on both Earth and Moon depends on the product of their masses. Also, both Earth and Moon exert the same force but in opposite directions.
Student 2 is wrong because the gravitational force is the only force acting between the Earth and Moon, and from Newton's third law, it follows an action-reaction pair. But, student 2 got it wrong in the sense that the magnitude of the action and reaction forces are the same and are equal to the gravitational force.
So, the gravitational force acting on each object is the same and doesn't vary for each mass.
We should agree with the student 1.
Gravitational force:
This is to be done due to the magnitude of the force since earth and moon should be based on the masses' product. Moreover, the earth and moon exert a similar force but that should be in inverse directions. Here student 2 should be wrong because the gravitational force is that force that acted between the earth and moon. Based on this, we can say that it should be acted on each object when it is the same also it should not be changed for every mass.
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I need help pls !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
figure d
Explanation:
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answer:
(a) Approximately [tex]0.335\; \rm m[/tex].
(b) Approximately [tex]1.86\; \rm m\cdot s^{-1}[/tex].
(c) Approximately [tex]0.707\; \rm m[/tex].
(d) Approximately [tex]0.228\; \rm m[/tex].
Explanation:
[tex]v_i[/tex] denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let [tex]m[/tex] denote the mass of the block. Let [tex]\mu[/tex] denote the constant of kinetic friction between the object and the surface. Let [tex]g[/tex] denote the constant of gravitational acceleration.Let [tex]k[/tex] denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy [tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2[/tex].
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of [tex]D[/tex] and compressed the spring by the same distance.
Energy lost to friction: [tex]\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D[/tex]. Elastic potential energy that the spring has gained: [tex]\displaystyle \frac{1}{2}\,k\, D^2[/tex].The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
[tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
Assume that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex]. In the equation above, all symbols other than [tex]D[/tex] have known values:
[tex]m =1.10\; \rm kg[/tex].[tex]v_i = 2.60\; \rm m \cdot s^{-1}[/tex].[tex]\mu = 0.250[/tex].[tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].[tex]k = 50.0\; \rm N \cdot m^{-1}[/tex].Substitute in the known values to obtain an equation for [tex]D[/tex] (where the unit of [tex]D\![/tex] is [tex]m[/tex].)
[tex]3.178 = 2.69775\, D + 25\, D^2[/tex].
[tex]2.69775\, D + 25\, D^2 + 3.178 = 0[/tex].
Simplify and solve for [tex]D[/tex]. Note that [tex]D > 0[/tex] because the energy lost to friction should be greater than zero.
[tex]D \approx 0.335\; \rm m[/tex].
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
[tex]\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J[/tex].
As the object moves to the left, part of that energy will be lost to friction:
[tex](\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J[/tex].
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
[tex]2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J[/tex].
Calculate the velocity corresponding to that kinetic energy:
[tex]\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}[/tex].
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ([tex]1.91\; \rm J[/tex]) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is [tex]\mu \cdot m \cdot g[/tex].
[tex]\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m[/tex].
(d)Similar to (a), solving (d) involves another quadratic equation about [tex]D[/tex].
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) [tex]1.91\; \rm J[/tex].
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
[tex]\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
[tex]25\, D^2 + 2.69775\, D - 1.90811\approx 0[/tex].
Again, [tex]D > 0[/tex] because the energy lost to friction is greater than zero.
[tex]D \approx 0.228\; \rm m[/tex].
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, [tex]v_i[/tex] = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
[tex]Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2[/tex]
Work done = Force × Distance
Friction force, [tex]F_f[/tex] = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
[tex]Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718[/tex]
Energy transferred by object = 3.718 J
[tex]Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2[/tex]
[tex]Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2[/tex]
[tex]W_{spring}[/tex] = 25·d²
Work done by friction, [tex]W_{friction}[/tex] = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, [tex]W_{friction}[/tex] = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
[tex]D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m[/tex]
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, [tex]W_{spring}[/tex] = 25·D'²
Work done on friction, [tex]W_{friction}[/tex] = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = [tex]W_{spring}[/tex] + [tex]W_{friction}[/tex]
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
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Imagine you are outside playing with a dog. The dog runs back and forth, left to right. What is the best way to describe how fast the dog's motion is?
The direction the dog moves.
The distance the dog moves over a certain amount of time.
The distance the dog moves.
Answer:
It is B.
Explanation:
This is because you want to see how fast the dog moves and the best way to find out is by seeing how fast it moves in a certain amount of time. pls brainly!
A boy pushes a box with a force of 150 N at an angle of 40 with a flat floor. What component of his force is directed downward , or into the floor . PLEASE ANSWER!!!!!
Answer:
[tex]F_y=96.4N[/tex]
Explanation:
Hello.
In this case, considering the force diagram shown on the attached picture, we can see that the component of his force is directed downwards is:
[tex]F_y=F\times sin (\theta)[/tex]
Because the other component is the horizontal one:
[tex]F_x=F\times cos(\theta)[/tex]
In this case, the y-component force turns out:
[tex]F_y=150N\times sin (40\°)\\\\F_y=96.4N[/tex]
Moreover, the x-component force is also computed if required:
[tex]F_x=150N\times cos(40\°)\\\\F_x=114.9N[/tex]
Best regards.
A metal wire has a circular cross section with radius 0.800 mm. You measure the resistivity of the wire in the following way: you connect one end of the wire to one terminal of a battery that has emf 12.0 V and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is d. You measure the current in the wire as a function of d. The currents are small, so the temperature change of the wire is very small. You plot your results as I versus 1/d and find that the data lie close to a straight line that has slope 600 A⋅m.
Required:
What is the resistivity of the material of which the wire is made?
Answer:
The value is [tex]\rho = 4.02 *10^{-8} \ \Omega \cdot m[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 0.800 mm = 0.0008 \ m[/tex]
The voltage of the battery is [tex]emf = 12.0 V[/tex]
The slope is [tex]s = 600 \ A \cdot m[/tex]
Generally the resistance is mathematically represented as
[tex]R = \frac{\rho * d }{A }[/tex]
Generally the current is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{V}{\frac{\rho * d }{A }}[/tex]
=> [tex]I = \frac{V * A }{\rho} * \frac{1}{d}[/tex]
Comparing this equation to that of a straight line we see that the slope is
[tex]s = \frac{V * A }{\rho}[/tex]
So [tex] 600 = \frac{V * A }{\rho}[/tex]
Here A is the cross-sectional area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.0008 )^2[/tex]
=> [tex]A = 2.011*10^{-6} \ m^2[/tex]
So
[tex] 600 = \frac{12.0 * (2.011*10^{-6}) }{\rho}[/tex]
=> [tex]\rho = \frac{12 * 2.011*10^{-6} }{600}[/tex]
=> [tex]\rho = 4.02 *10^{-8} \ \Omega \cdot m[/tex]
A car is traveling with a velocity of 20 M/S and has a weight (mass) of 1000 kg. The car has kinetic energy calculate it
Answer:
200000 joules or 200 kilojoules
Explanation:
Kinetic energy = 1/2m(v^v)
Kinetic energy = 1/2*1000* 20^20
Kinetic energy = 500*400
Kinetic energy = 200000 joules or 200 kilojoules
A 2070 kg space station orbits Earth at an altitude of 5.37×105 m. Find the magnitude of the force with which the space station attracts Earth. The ma
Answer:
9.9 × 10^-16N
Explanation:
Given that a 2070 kg space station orbits Earth at an altitude of 5.37×105 m. Find the magnitude of the force with which the space station attracts Earth.
Calculating the gravitational field strength.
g = Gm/R^2
Where G = 6.67408 × 10-11 m3 kg-1 s-2
g =(6.67408 × 10-11 × 2070)/5.37 × 10^5
g = 1.38069×10^-7 / 2.88369^11
g = 4.788×10^-19 m/s
Force F = mg
Substitute g into the formula
F = 2070 × 4.79×10^-19
F = 9.9 × 10^-16N
A 58.72 kg person holding a steel ball stands motionless on a frozen lake.
The person then throws the ball, which propels the person at 1.05 m/s to the
right and the ball 3.75 m/s to the left. If the initial momentum of the system is
zero, what is the mass of the steel ball?
Answer: 16.44 kg
Explanation:
Which shows the correct steps in the formation of an ionic bond between these atoms?
A magnesium atom accepts six electrons from the fluorine atoms → Each fluorine atom donates three of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom accepts two electrons from the fluorine atoms → Each fluorine atom donates one of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion
Answer:
D) A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion
Explanation:
sorry if its a bit late
The correct steps in the formation of an ionic bond are A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion. The correct option is D.
What are ionic bonds?The electrical attraction between two ions with opposing charges creates an ionic bond, also known as an electrovalent bond, in a chemical molecule.
Ionic compounds are created by ionic bonds, while covalent bonds create covalent compounds. Ionic bonds are created by a complete transfer of electrons, whereas covalent bonds are created by sharing electrons.
Covalent bonds are weaker than ionic ones. Ionic compounds have higher melting and boiling points.
Therefore, the correct option is D.
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Please help me with physics homework I just need help with C and D. Picture is included
Answer:
For C is B and for D is A
Explanation:
B because that's where the b car will do all the force to go up, and A because there the car doesn't have to do any force other than the natural, and it already brings that on it's own.
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3. A cat is moving at 18 m/s when it accelerates at 4 m/s? for 2 seconds. What is his new
velocity?
Answer:
26 m/s
Explanation:
v = u + at
Given that,
u(initial speed) = 18 m/sa (acceleration) = 4m/s^2t ( time) = 2sand v ( final velocity) = ?
v = 18 + ( 4 × 2)
v = 18 + 8
v = 26 m/s
If the maximum linear velocity her center of mass reaches is 2.3 m/s, what is her maximum angular velocity
Like linear velocity, angular velocity is the relationship between the variation of angular space at a given time.
It was agreed that the trajectory of the movement is clockwise. Thus, the movement that follows this orientation will be positive and the opposite, negative.
For the formula relating the linear velocity, it is necessary to take into consideration the following formula
[tex]v = \omega \times R[/tex]
For example, for a R = 0.738 m
[tex]2.3 m/s = \omega \times R \therefore\\ R = 0.738m \therefore\\\omega = 3.12 \frac{rad}{s}[/tex]
Helpppp guys plsss help !
Answer:
with what? I can help but with what
Answer:
2
Explanation:
a tractor with 2000 kg of mass pulls a plow equal to 80.0 kg, exerting a 200 N force on it. The tractor and plow set moves horizontally to the right with an acceleration of 0.500 m / s2. The resistance force that the soil exerts on the plow has a modulus, in Newton, equal to?
Answer:
The resistance force that the soil exerts on the plough is 160 Newtons.
Explanation:
Fr = m.a
Fr = 80. 0,5
Fr = 40 N
Ft - F = Fr
200 - F = 40
F = 200 - 40
F = 160 Newtons
Predict the energy transformation that occurs when Carson sands a piece of wood with sandpaper.
A. chemical energy to heat energy
B. heat energy to mechanical energy
C. mechanical energy to heat energy
D. mechanical energy to chemical energy
Answer:
C. Mechanical to heat i,m pretty sure
Explanation:
I took the test
Answer:
C or D
Explanation:
not really sure
water vapor is what kind of gas?
Answer:
the gaseous phase of water
Explanation:
"It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. "
hope this helps!!
Water vapor is a greenhouse gas.
I know this is a simple answer, but I sincerely hope this helped!
During a baseball game, a hitter strikes the ball with a bat. When this happens, the ball
and the bat exert a force on each other. Why does the ball accelerate away from the bat
more than the bat accelerates away from the ball?
a. The ball has less mass, so it exerts less force on the bat.
b. The bat exerts more force than the ball because the batter is exerting a force
on the bat as it hits the ball.
c. The ball has less mass, so the equal force on the ball and the bat causes
greater acceleration of the ball.
d. The ball has a greater velocity before the collision, so the force affects the ball
more than it affects the bat.
Answer:
The answer is c
Explanation:
The vehicle land-speed record has long been held by rocket cars. These vehicles resemble the high-speed rocket planes that were used in the early days of the space program, but they have heavy metal wheels. On September 5, 1997, the world land-speed record was set by the British-built Thrust SSC rocket car, which had a top recorded speed of 341 m/s. The kinetic energy of the car at this speed is 5.289 × 108 J. What is the car’s mass?
Answer:
9096.93 Kg
Explanation:
The following data were obtained from the question:
Velocity (v) = 341 m/s
Kinetic energy (KE) = 5.289×10⁸J.
Mass (m) =.?
The mass of the car can be obtained as follow:
KE = ½mv²
5.289×10⁸ = ½ × m × 341²
5.289×10⁸ = ½ × m × 116281
5.289×10⁸ = m × 58140.5
Divide both side by 58140.5
m = 5.289×10⁸ / 58140.5
m = 9096.93 Kg
Therefore, the mass of the car is 9096.93 Kg