During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.145 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 10.9 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball
Answer:
J = -0.522 m/s
Explanation:
Given that,
The mass of the baseball, m = 0.145 kg
Initial velocity, u = 14.5 m/s
Final velocity, v = 10.9 m/s
aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,
[tex]J=m(v-u)[/tex]
Substitute all the values,
[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]
The direction of impulse is opposite to the direction of velocity.
1 point
3. A 75 N box requires 250 J of work to move to a shelf. How high is the
shelf?
O 3.3 m
3.25m
0 3.9m
O 3.6m
Match each type of wave made during an earthquake to its wave form.
Answer:
I see anything so I can answer you .
What do you think will happen to the Lunar phases if the moon was hit by an asteroid?
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 6.9 s later. How high is the cliff?
In an elastic collision, ________. A. an individual molecule in the collision never loses energy B. the molecules involved in the collision move in a circular motion C. the total energy of all molecules in the collision remains constant D. the molecules involved in the collision are held to each other by strong intermolecular interactions
Answer:
Im pretty sure its B im very sorry if its wrong.
If we always drop the balls from 1-m
height in each trial, what type of variable
is this in this experiment?
Constant Variable
Dependent Variable
Independent Variable
Answer:
height
weight of ball
time of ball falling
Q1what is pinhole camera?
A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to two groups.students in group one get the tutorial support and those in group two do not.In this example what is dependent variable,independent variable,control groupand experimental group
The dependent variable is academic performance while the independent variable is presence/absence of tutorial support.
The correct results are:
The dependent variable is academic performanceThe independent variable is the presence/absence of tutorial supportThe control group are students who did not get the tutorial support.The experimental group were students that got the tutorial supportIn every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
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A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.
Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0
m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?
Answer:
v = 2.57 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. When it is at an angle of 30º
Em₀ = K + U = ½ m v₁² + m g y₁
final point. Lowest position
Em_f = K = ½ m v²
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v₁² + m g y₁ = ½ m v²
Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle
cos 30 = L ’/ L
L ’= L cos 30
y₁ = L -L '
y₁ = L- L cos 30
we substitute
½ m v₁² + m g L (1- cos 30) = ½ m v²
v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]
let's calculate
v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]
v = 2.57 m / s
A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?
Answer: North East
Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.
What is the only difference between the reactant and product side of a chemical reaction?
Answer:
Products is the result. Reactants produce the result
Explanation:
,,
Which of the following is true for the entropy of the universe?
A It is always decreasing.
B It is always increasing.
C It is always negative.
D It is always a constant.
Answer:
B It is always increasing.
Explanation:
In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.
This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.
The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.
Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.
Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.
1
Select the correct answer.
Which type of energy is thermal energy a form of?
A
chemical energy
B.
kinetic energy
C. magnetic energy
D. potential energy
Reset
Next
Answer:
B. kinetic energy
Explanation:
Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.
A 38.0 kg box initially at rest is pushed 4.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the change in kinetic energy of the box J (f) the final speed of the box m/s
Answer:
a) Wapp = 520 N
b) ΔUf = 447 N
c) Wn = 0
d) Wg = 0
e) ΔK = 73 J
f) v = 1.96 m/s
Explanation:
a)
Applying the definition of work, as the dot product between the applied force and the displacement, since both are parallel each other, the work done on the box by the applied force can be written as follows:[tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]
b)
The change in the internal energy due to the friction, is numerically equal to the work done by the force of friction.This work is just the product of the kinetic force of friction, times the displacement, times the cosine of the angle between them.As the friction force opposes to the displacement, we can find the work done by this force as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]
The kinetic force of friction, can be expressed as the product of the kinetic coefficient of friction times the normal force.If the surface is level, this normal force is equal to the weight of the object, so we can write (2), as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]
So, the increase in the internal energy in the box-floor system due to the friction, is -Wffr = 447 Jc)
Since the normal force (by definition) is normal to the surface, and the displacement is parallel to the surface, no work is done by the normal force.d)
Since the surface is level, the displacement is parallel to it, and the gravitational force is always downward, we conclude that no work is done by the gravitational force either.e)
The work-energy theorem says that the net work done on the object, must be equal to the change in kinetic energy.We have two forces causing net work, the applied force, and the friction force.So the change in kinetic energy must be equal to the sum of the work done by both forces, that we found in a) and b).So, we can write the following expression:[tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]
f)
Since the object starts at rest, the change in kinetic energy that we got in e) is just the value of the final kinetic energy.So, replacing in (4), we finally get:[tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]
Solving for v:[tex]v_{f} = \sqrt{\frac{2*\Delta K}{m} } = \sqrt{\frac{2*73J}{38.0kg}} = 1.96 m/s (6)[/tex]a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
What will be the work done?The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.
If the object achieve movement due to the work then the energy in the object will be kinetic energy.
If the object attains some height against the gravity then the energy in the object will be potential energy.
Now it is given in the question that
The horizontal force [tex]F_h=130\N[/tex]
mass of the object m= 38 kg
Coefficient of friction [tex]\mu=0.3[/tex]
Displacement of the object [tex]\delta x=4\ m[/tex]
(a) The work done will be
[tex]W=F_h\tines \Delta x[/tex]
[tex]W=130\times 4=520\ J[/tex]
(b) The increase in the internal energy
The increase in the internal energy of the box is due to the energy generated by the force of friction so
[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]
here [tex]F_f[/tex] is the frictional force and is given as
[tex]\mu=\dfrac{F_f}{R}[/tex]
Here R is the normal reaction and its value will be weight of the box in opposite direction.
[tex]\mu=\dfrac{F_f}{-mg}[/tex]
[tex]F_f=-mg\times \mu[/tex]
[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]
[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]
(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.
(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.
(e) The change in the KE of the box.
The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be
[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]
(f) The speed of the box
The KE of the box will be
[tex]KE=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]
Thus
a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
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Force exerted on a body changes it's
____ is a chemical messenger released by virus infected cells
Answer:
Cytokines is the answer
Explanation:
it is another word for the chemical messanger
9
os End
bilo
b) A battery of unknown emf is connected across resistances, as shown in fig below. The voltage
drop across the 8ohms resistor is 20V. What will be the current reading in the ammeter? What is the ammeter?what is the emf of the battery?
Answer is in the file below
tinyurl.com/wpazsebu
1. When an unbalanced force acts on an object,
O the object remains at rest
the weight of the object decreases.
O the object's motion changes.
the inertia of the object increases.
Which of the following is something you could do to reduce your impact on the environment?
Throw newspapers in the trash can.
Drive a car instead of walking short distances.
Open a window on a cold day while the heat is on.
Take shorter showers.
Answer:
Take shorter showers.
Throw newspapers in the trash can.
Name one similarity and one difference between a set and a bump in volleyball??
One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.
One similarity and one differenceOne similarity between a set and a bump in volleyball is the movement and use of legs and hands.
Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand, the basic pass in volleyball is known as bump.
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Rhodium is in period 5 of the periodic table. What does this tell you about this element
(It’s number 4 in the picture)
Rhodium (Rh) is in period 5 on the periodic table. What does this tell you about this element ?
option - D
It has 5 energy levels/shells in its electron cloud.
With the information from the periodic table we can find that the correct statement is:
D. it is 5 energy level
The periodic table of the elements is a method of classifying the chemical elements based on their atomic number, it is divided into rows and columns
The rows are called the period and indicate the energy shell that is filled with electrons, all periods start with one electron and end when the shell is filled with 8 electrons.
The columns called groups correspond to elements that have the same number of electrons in their last shell, the chemical properties of the elements of the group are similar.
In this case we have an element the Rhodium, from period 5 and group VIIIB, when reviewing the period table it can be indicated that it has properties similar to cobalt and the inner layer is filled with two electors in its last layer.
When reviewing the different claims
a) False. The periodic table is not about molecules but about atoms
b) False. The metallic characteristics depend on the relationship between the number of electrons and the fermi level that is not indicated in the periodic table.
c) False. The number of protons is equal to the atomic number, in this case the rhodium has Z = 45, therefore it has 45 protons in the nucleus
d) True. It is at level 5 since the energy levels and the periods coincide.
In conclusion, with the information from the periodic table, we can find that the correct statement is:
D. it is 5 energy level
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.00 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) 1.2 m/s
b) 0.350 m
c) 1.2 m/s
d) 0.250 m
Explanation:
a)
At any traveling wave, there exists a fixed relationship between the frequency, the speed and the wavelength, as follows:[tex]v = \lambda * f (1)[/tex]
We know that the frequency is equal to the number of cycles that the wave does in a second, so it's the inverse of the time needed to complete one cycle, that we call the period.In our case, since the wave completes one half cycle (from the highest point to the lowest one) in 2.60 s, this means that the time needed to complete a cycle, is just the double of it, i.e., 5.20 s .The frequency of the wave is just the inverse of this value:⇒ f = 1/T = 1/5.20 s = 0.2 1/s = 0.2 Hz (2)If the wave crests are spaced a horizontal distance of 6.0 m apart, this means that the wavelength λ is just 6.0 m.Replacing in (1) by (2) and the given λ, we can find the speed v, as follows:[tex]v = \lambda * f = 6.0 m * 0.2 1/s = 1.2 m/s (3)[/tex]
b)
By definition, the amplitude of the wave, is the absolute value of its highest value over its the zero reference level (in this case the surface of the water), so it is the half of the total vertical distance traveled by the boat:⇒ A = Δd/2 = 0.700m/2 = 0.350 m (4)
c)
Since the amplitude of the wave is independent of the frequency and the wavelength (that define the speed of the wave) the wave speed remains the same, i.e., 1.2 m/s.d)
If the total distance traveled by the boat were 0.500 m , the amplitude would be just half this value, as follows:⇒ A = Δd/2 = 0.500m/2 = 0.250 m (5)
Sound travels at a speed of 330 meters/second. If Denise hears a police siren 150 meters away, approximately how long did it take for the siren sound to
travel from the police vehicle to her?
Answer:
It went about 2 meters away
Explanation:
Diffraction occurs for all types of waves, including sound waves.
a. True
b. False
Answer:
a. True
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
[tex] Speed = wavelength * frequency [/tex]
Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.
Diffraction occurs for all types of waves, including sound waves.
prove that d1=R(d1-d2) in relative density
A car with a mass of 1200 kg has a momentum of 15, 350 kg * m/s. What is its velocity?
Answer:
v = 12.79 m/s
Explanation:
Given that,
The mass of a car, m = 1200 kg
Momentum of the car, p = 15 350 kg-m/s
We need to find the velocity of the car. We know that, the formula for the momentum of an object is given by :
p = mv
Where
v is the velocity of the bject
So,
[tex]v=\dfrac{p}{m}\\\\v=\dfrac{15350}{1200}\\\\v=12.79\ m/s[/tex]
So, the velocity of the car is 12.79 m/s.
A ball rolls along the floor with a constant velocity of 12 m/s. How far will it have
gone after 6 seconds?
Answer:
72m
12m x 6s=72m
there ya go
An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
