Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
(A) Electricity and Magnetism
A). Three point charges are aligned along the x axis as shown in
Fig. Find the electric field at (a) the position (2, 0) and (b) the
position (0, 2).
electricity
Explanation:
the position (2,o
How much would a 15.0 kg object weigh on that planet? Round the answer to the nearest whole number.
Answer:
168
Explanation:
Answer: a 15 kg object would weigh the most on Neptune
168 N
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
a motor boat is traveling at 25 knots towards 340 degree on a river flowing at 20 knots towards 175 degrees. What is the actual speed of the boat as seen by a helicopter piolet hovering above?
Answer:
Vbx = 25 * cos 340 = 23.5 knot x-component of boats speed
Vrx = 20 cos 175 = -19.9 knot x-component of rivers speed
Vx = 3.58 knot x-component of boat and river speed
Vby = 25 sin 340 = -8.55 knot y-component of boat speed
Vry = 20 sin 175 = 1.74 knot y-component of river speed
Vy = -6.81 knot y-component of boat and river speed
V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots
Can we cure aging? Support your answer?
Answer: yes it can
Explanation:
Answer:
Yes you can. You need to eat healthy and have the proper diet and always exercise and build muscles, this can slow down aging, and most importantly, we must always be calm. However there is no way to stop aging for that’s part of life.
Explanation:
The light bulbs are identical. Initially both bulbs are glowing. What happens when the switch is closed
Answer:
They turn off
Explanation:
A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming free fall conditions, what was the rocket's initial upward velocity?
a. 98.0 m/s
b. 123 m/s
c. 24.5 m/s
d. 49.0 m/s
d. 49.0 m/s
this is your answer. ....OK. ..
please mark me as brainliest. .....
follow me.....my friend. .....
1. A hot air balloon weighing 30 N is tied to the ground by a string to prevent
from floating off the ground. The volume of the balloon is 20 m and the
density of air is 1.3 kgm ?. Find:
a. upthrust acting on the balloon. Take g = 10 ms 2.
b. force exerted by the rope on the balloon?
Answer:
a) FB = 260 [N]
b) FT = 230 [N]
Explanation:
In order to solve this problem we must use a static analysis, since Globe does not move. For a better understanding in solving this problem, a free body diagram with the forces acting on the globe is attached.
The buoyant force acts upward as it causes the balloon to tend to float, the weight of the balloon tends to lower the balloon and the downward tension force does not allow the balloon to float
The buoyant force is defined by the following equation:
FB = Ro*V*g
where:
FB = Buoyant force [N]
Ro = density of the air = 1.3 [kg/m^3]
V = volume of the balloon = 20 [m^3]
g = gravity acceleration = 10 [m/s^2]
FB = 1.3*20*10 = 260 [N]
Now we do a sum of forces equal to zero in the y-axis
FB - 30 - FT = 0
260 - 30 = FT
FT = 230 [N]
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
Learn more about "Friction" here:
brainly.com/question/13357196
21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?
Answer:
v = 66 m/s
Explanation:
Given that,
The initial velocity of a car, u = 0
Acceleration of the car, a = 11 m/s²
We need to find the final velocity of the toy after 6 seconds.
Let v is the final velocity. It can be calculated using first equation of motion. It is given by :
v = u +at
v = 0 + 11 m/s² × 6 s
v = 66 m/s
So, the final velocity of the car is 66 m/s.
Which of these statements best describes the impact of ocean thermal power and current power on the environment?
A. Current power may decrease the fish population.
B. Current power may decrease the gravitational pull of the moon.
C. Ocean thermal power may increase the fish population.
C. Ocean thermal power may increase the gravitational pull of the moon.
Answer:
A. Current power may decrease the fish population.
Explanation:
The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.
The environment is made up of living and non-living components that co-exist and interact with one another.
Harnessing current power from ocean movement will seriously affect the fish population. Most fishes are not sedentary. They move and glide through the water. When current power causes a change in the environment of the fish. This will definitely affect the normal condition prevalent in the body of water.Answer:
A
Explanation:
I took the test good luck :D
-. What is the acceleration of 4 kg trolling bag pulled by a girl with a
force of 3 N?
Answer:
Acceleration(a) = 0.75 m/s²
Explanation:
Given:
Force(F) = 3 N
Mass of thing(m) = 4 kg
Find:
Acceleration(a)
Computation:
Force(F) = ma
3 = (4)(a)
Acceleration(a) = 3/4
Acceleration(a) = 0.75 m/s²
A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.
Answer:
xn = 36.28 [m]
Explanation:
To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.
[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]
where:
x - xo = displacement of the car [m]
Vo = initial velocity = 0
t = time = 7.2 [s]
a = acceleration = 1.4 [m/s^2]
The initial velocity is zero, as the car begins its movement from rest.
xn = (x - xo), Now replacing
xn = (0*7.2) + 0.5*1.4*(7.2^2)
xn = 36.28 [m]
Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD.
Answer:
KE= 1/2mv²
Explanation:
The kinetic energy of a body is the energy possessed by virtue of the body in motion
Given the parameters
m which is the mass of the body
v which is the velocity of the body too
K.E = kinetic energy
The expression for the kinetic energy of a body is given as
KE= 1/2mv²
a Venus fly trap is an......... plant.
Answer:
A Venus flytrap is a perennial carnivorous plant.
I hope it helps
plant that eats meat
Describe four ways in which improving your muscular fitness would have a positive affect in your life
Four ways in which improving our muscular fitness would have a positive affect on our lives are given below:
Injury preventionBody compositionChronic disease prevention.Lifetime muscle and bone healthMuscular fitnessMuscle fitness simply means having muscles that can lift heavier objects or muscles that will work longer before becoming exhausted.
It helps us improves when a person does activities that build or maintain muscles or that increase how long a person can use his or her muscles.
Learn more about muscular fitness:
https://brainly.com/question/2415901
A car takes 8 s to increase its velocity from 10 m/s to
30 m/s. What is its average acceration?
Answer:
Average acceleration is 2.5 m/s^2.
Explanation:
Average acceleration can be found by dividing the change in speed by the change in time.
We know that the initial velocity=10m/s, the final velocity=30m/s and time elapsed is t=8s.
a= (Final velocity-Initial Velocty)/t
a=(30-10)/8=20/8=2.5 m/s^2
An increase in temperature the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container . Answer Bank What temperature must a gas, initially at 10 ∘C, be brought to for the pressure to triple?
Answer:
a
The pressure will increase
b
[tex]T_2 = 576^oC[/tex]
Explanation:
From the ideal gas law we have that
[tex]PV = nRT[/tex]
We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase
The initial temperature is [tex]T_i = 10^oC = 10 + 273 = 283 \ K [/tex]
The objective of this solution is to obtain the temperature of the gas where the pressure is tripled
Now from the above equation given that nR and V are constant we have that
[tex]\frac{P}{T} = constant[/tex]
=> [tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
Let assume the initial pressure is [tex]P_1 = 1 Pa[/tex]
So tripling it will result to the pressure being [tex]P_2 = 3 Pa[/tex]
So
[tex]\frac{1}{283} =\frac{3}{T_2}[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 849 \ K [/tex]
Converting back to [tex]^oC[/tex]
[tex]T_2 = 849 - 273[/tex]
=> [tex]T_2 = 576^oC[/tex]
Question 16 1 pts Jessie feels pressured by his parents to get a job. This is an example of the law of?
readiness
disuse
effect
belonging
A man walking at 3.56 m/s accelerates at 2.50 m/s2 for 9.28 s. How far does he get?
141 m
26.8 m
44.6 m
248 m
50 POINTS!!! PLEASE!!!!!! Pls help me with this been stuck on it in like FOREVER!! PLSS.
Answer:
it will move in 356.5g in diagonal direction(resultant or between the both direction)
Explanation:
285+428/2=356.5g
Hope u like it
PLZ MARK ME AS BRAINLIEST
2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she 4covers its 48.0 m length at a speed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0 m length at a speed of 5.00 m/s. How long does it take Suzette to make to class? Did Suzette beat the bell?
Answer:
62 secondsnoExplanation:
The total travel time Suzette experiences is the sum of the times in each hallway. Using
time = distance/speed
we can add the times.
(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)
= 10 s + 40 s + 12 s
= 62 s
It takes Suzette 62 seconds to get to class. She does not beat the bell.
Please help me! What is Ohm's law?
Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:
voltage = current x resistance
This formula tells you that current and resistance is the voltage of an energy bond.
Hope this helps you!
Answer: What is Ohm’s Law?
Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.
E = I × R
It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.
Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.
Plz mark brainliest:)
The world’s fastest car can accelerate from rest to 60 mph(27m/s) in 2.2 seconds. What is the magnitude of its acceleration?
Hi there! :)
12.27 m/s²Find the acceleration by finding the change in velocity over time:
(vf - vi) / (time) = acceleration over the interval
*Remember, an object at rest has a velocity of 0 mph, and use the correct units for speed that would result in an acceleration involving meters / seconds.*
(27 - 0) / (2.2) = 27 / 2.2 ≈ 12.27 m/s²
The speed of a car is decreasing from 35 m/s to 15 m/s in 4s
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
Open Box. Consider a hollow box with the top missing. The sides have negligible thickness and each has length L and mass m. (a) Find the x-coordinate of the center of mass.
Answer:
x_{cm} = L / 2
Explanation:
The center of mass is defined by
[tex]x_{cm}[/tex] = 1 / M ∑ m_{i} x_{i}
where M is the total mass of the system
in this case the system is continuous so, for which we use the density
ρ = dm / dx
dm = ρ dx
substituting
x_{cm} = 1 / M ∫ x ρ dx
x_{cm} = ρ / M ∫ x dx
we integrate and evaluate from x = 0 to x = L
x_{cm} = ρ / M (L² /2 -0)
we introduce the density which is constant
ρ = M / L
x_{cm} = 1 /M (M/L) L² / 2
x_{cm} = L / 2
A curve that has a radius of is banked at an angle of . If a -kg car navigates the curve at without skidding, what is the minimum coefficient of static friction between the pavement and the tires
Answer:
0.65
Explanation:
For whatever reasons, the parameters are not giving. So, I will assume by myself to make the calculations easier. You can substitute whatever it is to it from your question.
Given that
Radius of the road, r = 63 m
Speed of the car, v = 20 m/s
The relationship between a car that is passing through a curve and it's frictional force is said to be
U(s) * g = v²/r
In the formula above,
U(s) = coefficient of static friction
g = acceleration due to gravity, 9.8 m/s²
v = velocity of the car
r = radius of the road
Now when we substitute the earlier stated values, we have
U(s) * 9.8 = 20² / 63
U(s) * 9.8 = 400 / 63
U(s) * 9.8 = 6.35
U(s) = 6.35 / 9.8
U(s) = 0.65
Thus, our coefficient of static friction, based on the stated values is 0.65
Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.
Answer:
F=1.3x10^-2N
Explanation:
Fe= k(6x10^-7C)^2/(0.5)^2
Electrical force of attraction between the balloons is F=1.3x10^-2N
The electric force of attraction between two balloons should be F=1.3x10^-2N.
Calculation of the electric force;Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.
So, here the electric force is
Fe= k(6x10^-7C)^2/(0.5)^2
F=1.3x10^-2N
hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.
Learn more about force here: https://brainly.com/question/19848845
What does the principle of superposition help scientists determine?
A) The super powers of a rock layer
B) The exact and absolute age of a rock layer
C) The relative age of a rock layer
D) The position of a fossil
Answer:
B
Explanation:
the exact and absolute age of a rock layer
Answer:
The relative age of a rock layer.
Explanation:
The answer is C.