During a football workout, two linemen are pushing the coach on the sled. The
combined mass of the sled and the coach is 300 kg. The coefficient of friction between
the sled and the grass is 0.800. The sled accelerates at the rate of 0.580 m/s^2.
Determine the force applied to the sled by the lineman.
Answer:
F_{players} = 2528.4[N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body must be equal to the product of mass by acceleration.
On the sled act two forces, the force of friction and The Force executed by the football players.
The movement of the football players will be taken as positive, while the friction will be taken as negative, since it is opposed to the movement of the sled.
ΣF = m*a
where:
F = force [N]
m = mass [kg]
a = acceleration = 0.580 [m/s²]
[tex]F_{players} - f_{friction} = m*a[/tex]
The friction force is defected as the coefficient of friction by the normal force, the normal force on a horizontal surface can be calculated as the product of mass by gravitational acceleration.
[tex]f_{friction}=0.8*(300*9.81)\\f_{friction}=2354.4[N][/tex]
Now we can calculate the force exerted by the players.
[tex]F_{players}-2354.5=300*0.580\\F_{players}=2528.4[N][/tex]
Block A of 15kg of the figure below slides on a surface where the coefficient of kinetic friction is uk=0.3. The block moves at 10m/s when it is s=4m from a second 10kg B block, initially at rest. Considering this situation, determine what will be the maximum compression of the spring, initially decompressed, after the collision if the coefficient of return between the blocks is e=0.6. Tip here, as the blocks are subject to friction forces, use the equation that relates mechanical energy and the work done by non-conservative forces.
Answer:
0.287 m
Explanation:
The velocity of block A when it reaches block B is:
KE₀ = KE + W
½ mv₀² = ½ mv² + Fd
½ mv₀² = ½ mv² + mg μ d
v₀² = v² + 2g μ d
v² = v₀² − 2g μ d
v² = (10 m/s)² − 2 (9.8 m/s²) (0.3) (4 m)
v = 8.75 m/s
The coefficient of restitution is 0.6, so the relative velocity between A and B after the collision is:
e = |Δv after| / |Δv before|
0.6 = Δv / (8.75 m/s)
Δv = 5.25 m/s
Momentum is conserved, so the speed of block B after the collision is:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
(15 kg) (8.75 m/s) + (10 kg) (0 m/s) = (15 kg) (v) + (10 kg) (v + 5.25 m/s)
131.2 m/s = 25v + 52.5 m/s
25v = 78.7 m/s
v = 3.15 m/s
Energy is conserved, so the compression of the spring is:
KE = EE + W
½ mv² = ½ kd² + Fd
½ mv² = ½ kd² + mg μ d
½ (10 kg) (3.15 m/s)² = ½ (1000 N/m) d² + (10 kg) (9.8 m/s²) (0.3) d
49.6 = 500 d² + 29.4 d
0 = 500 d² + 29.4 d − 49.6
d = [ -29.4 ± √(29.4² − 4(500)(-49.6)) ] / 1000
d = (-29.4 ± 316.2) / 1000
d = 0.287 m
The difference between a law and a theory is the diffrence between what and why . Explain
Answer:
A law is defined as a description or a statement given after an observed phenomenon. A theory is a simplification of certain observational data as to how and why it happened.
Explanation:
I hope this helped solve your question.
What is the relationship between lightning and atoms?
A. Electrons colliding in the air create lightning
B. The nucleus of atoms colliding creates an atom
C. Lightning is made of atoms
D. Lightning is created by the collision of protons
Answer:
C. Lighting is made of atoms
Explanation:
Everything is made of atoms that are, in turn, made out of charged particles. All charged particles come in one of two types: positive and negative (or plus and minus). The minus particles are the electrons, and the plus particles are the much heavier protons which are buried deep in the nucleus.
The frequency of an electromagnetic wave is 1.0 x 105 Hz. Calculate the wavelength in meters. Show your work,
and be sure to keep track of the units.
Answer:
wavelength = 3000 m
Explanation:
We use the formula for the relationship between the speed (c), frequency (v), and wavelength [tex](\lambda)[/tex]:
[tex]c=\lambda\,*\,\nu\\300000000 \,\,m/s = \lambda\,*100000 \,\,1/s\\\lambda \,=\,3000\,\,m[/tex]
A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?
If the sound source is moving then the pitch of the sound will ___________________ to the observer.
Answer:
Increase
Explanation:
The frequency of sound determines the sound. If the frequency is lower the pitch will be lower. If the frequency is higher the pitch will be higher. This is affected by the motion of the sound source because when a sound source is moving faster the frequency will become higher.
A red blood cell contains 4.8 107 free electrons. What is the total charge of these electrons in the red blood cell?
Answer:
Charge, [tex]q=7.68\times 10^{-12}\ C[/tex]
Explanation:
It is given that,
The number of electron in a RBCs, [tex]n=4.8\times 10^7[/tex]
We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :
q = ne
e is the change on electron
[tex]q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C[/tex]
So, the net charge is [tex]7.68\times 10^{-12}\ C[/tex].
It the same child has a velocity of 2 m/s half-way down the slide, what is his kinetic energy?
Answer:
[tex] \frac{1}{2} \times m \times {v}^{2} \\ \frac{1}{2} \times m \times {2}^{2} \\ 2 \times m[/tex]
I dont the child mass...you should substitute that value to (m)
then you can get your answer
Which type of force is the weakest?
The correct option is A i.e. Van der Waals forces.
what is Van der Waals forces?
Van der Waals forces are weak intermolecular forces that are depending on atom or molecule distance. Interactions between uncharged atoms/molecules produce these forces.
The variation in the polarizations of two particles close to each other can produce Van der Waals forces.Van der Waals forces are substantially weaker than covalent and ionic bonding.The Van der Waals forces are additive in nature, consisting of multiple distinct interactions.These forces are inexhaustible.These forces have no discernible directional characteristics.They aren't affected by temperature (except dipole-dipole interactions)Short-range forces are Van der Waals forces. When the atoms/molecules in question are near together, their magnitude is large.To know more about Van der Waals forces here
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A cannonball is launched diagonally with an initial horizontal speed of 51.0m/s
and an initial vertical speed of 24.0m/s. Label the hypotenuse, opposite side and
adjacent side, and determine all unknowns.
m/s
initial speed:
initial horizontal speed:
initial vertical speed:
m/s
m/s
At what angle () was the cannonball launched?
• initial horizontal speed: 51.0 m/s
• initial vertical speed: 24.0 m/s
• initial speed: √((51.0 m/s)² + (24.0 m/s)²) ≈ 56.4 m/s
• angle: tan(θ) = (24.0 m/s) / (51.0 m/s) → θ ≈ 25.2º
Using Pythagoras, the initial speed and the launch angle of the cannon ball are 56.36 m/s and 25.20° respectively.
The initial speed can be obtained using length of the diagonal :
v² = 51² + 24²
v² = 2025
v = √3177
v = 56.36 m/s
The launch angle, θ :
Tanθ = opposite / Adjacentθ = tan¯¹(24/51)
θ = tan¯¹(0.4705)
θ = 25.20°
Hence, the angle of launch is 25.20°
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A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom o the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is 0.02.
a. What is the kinetic energy of the block at the bottom of the curved surface?
b. What is the speed of the block at the bottom of the curved surface?
c. Find the stopping distance of the block?
d. Find the elapsed time of the block while it is moving on the horizontal part of the track.
e. How much work is done by the friction force on the block on the horizontal part of the track?
Answer:
a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J
Explanation:
a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.
So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.
So, mgh + 0 = 0 + K'
K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m
So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J
b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve
So, v = √(2K/m)
= √(2 × 186.2 J/5 kg)
= √(372.4 J/5 kg)
= √(74.48 J/kg)
= 8.63 m/s
c. To find the stopping distance, from work-kinetic energy principles,
work done by friction = kinetic energy change of block.
So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance
ΔK = -fd
K" - K' = - μmgd
d = -(K" - K')/μmg
Substituting the values of the variables, we have
d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)
d = -(- 186.2 J)/(0.98 kg m/s²)
d = 190 m
d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m
So, a = (v² - u²)/2d
substituting the values of the variables, we have
a = (0² - (8.63 m/s)²)/(2 × 190 m)
a = -74.4769 m²/s²/380 m
a = -0.2 m/s²
Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.
t = (v - u)/a
t =(0 m/s - 8.63 m/s)/-0.2 m/s²
t = - 8.63 m/s/-0.2 m/s²
t = 43.2 s
e. The work done by friction W = fd where
= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m
W = 0.02 × 5 kg × 9.8 m/s² × 190 m
W = 186.2 J
The potential energy of the loss of the block will be equal to the kinetic energy gain. The kinetic energy of the block is 186.2 J at the bottom of the curved surface.
The potential energy of the loss of the block will be equal to the kinetic energy gain.
So,
[tex]U = mgh[/tex]
Where,
[tex]U[/tex] - potential energy
[tex]m[/tex] - mass of block = 5 kg
[tex]g[/tex] - gravitational acceleration = 9.8 m/s²
[tex]h[/tex] = height = radius of curve = 3.8 m
Put the values in the formula,
[tex]U = 5 \times 9.8 \times 3.8 \\\\ U = 186.2 \rm \ J[/tex]
Therefore, the kinetic energy of the block is 186.2 J at the bottom of the curved surface.
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Why do I feel so sad? I'm not sure if its depression, because i will be fine not happy, but fine all day than randomly in the day I will start to feel sad. I've tried to take my life twice before the past few monthes
You are most likely going through depression. I think you should talk to someone, it could help or try talking to one of us on brainly. I hope you feel better eventually. Your life is valuable try and realize that.
Answer:
You don't need a reason to be sad. It sound as though you are suffering from depression. I understand that. Last year I thought about taking my life, but my parents found out how I was feeling and have been so supportive. Something My mom said that really got to me was, if you killed yourself, who would find you. For me it would have been my little sister. I would have scared her for life. I have come a long way since then, I am doing family therapy, as well as persona, therapy. If you are feeling sad, there is a hotline you can text rather than call. Let them talk you out of it. They won't call anyone if you don't ask them to. Talking to a complete stranger helps, they don't know anything about you so they can't judge you. Also try listening to music, find an artist that you can relate to. For me I like listening to NF when I am feeling down. It helps.
An ideal transformer has 60 turns in its primary coil and 360 turns in its secondary coil. If the input rms voltage for the 60-turn coil is 120 V, what is the output rms voltage of the secondary coil
a. 240 V
b. 720 V
c. 360 V
d. 480 V
e. 20 V
Answer:
720 V
Explanation:
Given that,
The number of turns in primary coil, N₁ = 60
The number of turns in secondary coil, N₂ = 360
The input rms voltage, V₁ = 120 V
We need to find the output rms voltage of the secondary coil . The relation between number of turns in primary coil - secondary coil to the input rms voltage to the output rms voltage is given by :
[tex]\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}\\\\V_2=\dfrac{N_2V_1}{N_2}\\\\V_2=\dfrac{360\times 120}{60}\\\\V_2=720\ V[/tex]
So, the output rms voltage of the secondary coil is 720 V. Hence, the correct option is (b).A 150.0 kg cart rides down a set of tracks on four solid steel wheels, each with radius 20.0 cm and mass 45.0 kg. The tracks slope downward at an angle of 17 ∘ to the horizontal. If the cart is released from rest a distance of 27.0 m from the bottom of the track (measured along the slope), how fast will it be moving when it reaches the bottom
Answer:
13.4 m/sExplanation:
given
Mass of cart= 150kg
mass of each wheel=45kg
mass of 4 wheels= 180kg
angle of the track= 17 ∘
distance of track= 27m
The height of the tracl is calculated thus:
sin 17° = h / 27
h = sin 17*27
h=7.89m
"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "
1. Potential energy at top= (M+4m)gh
2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²
3. rotational energy of wheels at bottom= 4(1/2 Iω²)
The total is expressed as
(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1
we know that I = mr² / 2
Put I= mr² / 2
(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)
(M+4m)gh = 1/2 (M+4m)v² + m r² ω²
we know that v²= r² ω²
(M+4m)gh = 1/2 (M+4m)v² + m v²
(M+4m)gh = v² (M/2 + 2m + m)
(M+4m)gh = v² (M/2 + 3m)
v = √[(M+4m)gh / (3m + M/2)]
v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]
v=√25516.26/142.5
v=√179.06
v = 13.4 m/s
An airplane with a mass of 5,000 kg needs to accelerate 5 m/s2 to take off before it reaches the end of the runway. How much force is needed from the engine
Answer:
25000 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 5000 × 5
We have the final answer as
25000 NHope this helps you
The towel has better blank
than that towel.
A.duplication
B.formation
C.qualification
D.absorption
Answer:
D
Explanation:
Towels Absorb
1) What do (x) and (y) symbolize?
a) What equations belong to each one?
b) How does horizontal motion effect vertical motion and vis versa?
c) What do both horizontal and vertical motion use?
Answer:
x is vertical and y is horizontal
Explanation:
A Carnot engine takes in heat at a temperature of 790 K and releases heat to a reservoir at a temperature of 470 K. What is its efficiency?
Answer: 1280 kilometers :D
Explanation:
790 K ' 470 K = 1280 KM (KILOMETERS)
Answer:
GIVE DIS GUY BRAINLIST
Explanation:
What did Rutherford conclude from the motion of the particles when shot through a thin gold foil
Answer:
He concluded the fact that most alpha particles went straight through the foil is evidence for the atom being mostly empty space. A small number of alpha particles being deflected at large angles suggested that there is a concentration of positive charge in the atom.
Explanation:
hii i need these answers please!! as soon as possible and if you can’t see anything jus let me know and ill tell you what it says
Answer:
Explanation:
CANT SEE ITS TOO PIXELATED
What is the mass, in kg, of a 136 pound gymnast on Earth?
Answer:
61.6886 kg
Explanation:
Answer:
61 kg
Explanation:
there ya go hope this was useful
Which physical activity is NOT aerobic exercise?
A) Hip-hop dancing
B) Jump roping
C) Yoga
D) Jogging
Answer:
C
Explanation:
Answer:
Hip-hop Dancing
Explanation:
Aerobic exercises are light (somewhat, a more accurate word would be bearable) workouts that you can endure over long periods of time. Anaerobic workouts describe workouts that require sudden burst of energy, like the ones you seen in most forms of dance, such as Hip-hop dance.
Hope this helps!
An object weighs 3 lb. at 10 earth radii from its center. What is the object's weight on the earth's surface?
______ lb.
300
30
3
3,000
Answer:
3
Explanation:
anybody wanna play among us?
Answer:
yess
Explanation:
Answer:
yes me
mark me brainliestAn engineer measures the velocity of a remote-controlled cart on a straight track at regular time intervals. The data are shown in the graph above. During which of the following time intervals did the cart return to its position at time t=0 s?
A. 7 s ≤ t < 10 s
B. 10 s ≤ t < 12 s
C. 5 s ≤ t < 6 s
D. 3 s ≤ t < 5 s
Answer:
A. 7 s ≤ t < 10 s
Explanation:
From Definite Integral Theory and Mechanical Physics we remember that change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:
[tex]A_{1}+A_{2} = 0[/tex] (1)
Where:
[tex]A_{1}[/tex] - Change in position from [tex]t = 0\,s[/tex] and [tex]t = 3\,s[/tex], measured in meters.
[tex]A_{2}[/tex] - Change in position from [tex]t = 5\,s[/tex] and [tex]t = T[/tex], measured in meters.
By definitions of triangle, we expand the equation above:
[tex]\frac{1}{2}\cdot (3\,s)\cdot \left(1.2\,\frac{m}{s} \right) -\frac{1}{2}\cdot (T-5\,s)\cdot v = 0\,m[/tex]
[tex]1.8\,m-0.5\cdot T \cdot v +2.5\cdot v = 0[/tex]
And the time is now cleared:
[tex]1.8\,m+2.5\cdot v = 0.5\cdot T\cdot v[/tex]
[tex]T = \frac{1.8+2.5\cdot v}{0.5\cdot v}[/tex]
Where:
[tex]v[/tex] - Final velocity of the cart, measured in meters per second.
[tex]T[/tex] - Time, measured in seconds.
If we know that [tex]v = 1\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1)}{0.5\cdot (1)}[/tex]
[tex]T = 8.6\,s[/tex]
The value does not coincide with the time from the graph.
If we know that [tex]v = 1.5\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1.5)}{0.5\cdot (1.5)}[/tex]
[tex]T = 7.4\,s[/tex]
This value does not coincide with the time from the graph.
If we know that [tex]v = 1.1\,\frac{m}{s}[/tex], then the intended instant is:
[tex]T = \frac{1.8+2.5\cdot (1.1)}{0.5\cdot (1.1)}[/tex]
[tex]T = 8.273\,s[/tex]
This results offer a reasonable approximation, which that the correct answer is A.
The car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.
The change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:
A + A' = 0 ..................................................(1)
A is the change in position from t = 0 and t = 3 s, measured in metres.
A' is the change in position from t = 5 and t = T s, measured in metres.
As per the triangle law,
1/2 ×(3 s)(1.2) - 1/2 ×(T - 5)v = 0
1.8 - 0.5T(v) +2.5(v) =0
T = (1.8 + 2.5v)/ (0.5v)
Here, v is the final velocity of cart.
For v = 1 m/s, we have,
T = (1.8 + 2.5(1))/ (0.5(1))
T = 8.6 s
Since, the values do not coincide with the time from the graph, so taking another value, v = 1.5 m/s to obtain the time as,
T = (1.8 + 2.5(1.5))/ (0.5(1.5))
T = 7.4 s
This value does not coincide with the time from the graph. If we know that , v = 1.1 m/s, then the intended instant is:
T = (1.8 + 2.5(1.1))/ (0.5(1.1))
T = 8.273 s
Thus, we can conclude that the car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.
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A cannonball is fired straight up at a speed of 25 m/s. What is the maximum altitude that it will reach?
Answer:
When the projectile is launched straight up, there isn't a horizontal ... The initial acceleration was 9.8 m/s2 pointing up, so the acceleration at any other point should be the same.
Explanation:
Hope it helped =)
The maximum altitude that the cannonball will reach if fired straight up at a speed of 25 m/s is; 31.86 m
According to the question;
The cannonball is fired straight up at a speed of 25 m/s
Additionally, the cannonball is fired against the force of gravity.
Consequently, the motion is in the opposite direction of the acceleration due to gravity.
From the equation of motion;
V² = U² - 2gHAt the maximum altitude, V = 0.
0² = 25² - (2× 9.81) H19.62H = 625H = 625/19.62H = 31.86mThe maximum altitude that it will reach is;
H = 31.86mRead more:
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1) An object travels 15 m in 3 s. What is its' speed?
Answer:
5m/s
Explanation:
Distance/Time= Speed
D/T=S
15/3=S
5=S
A student trying to calculate the coefficient of friction between a box and a surface. She measures that the 80kg box will slide if the student pushes with a force greater than 400n.
Answer:
[tex]\mu_s=0.51[/tex]
Explanation:
Coefficient of friction
The static coefficient of friction is a measure of the force needed to start moving an object from rest along a rough surface.
It's calculated as:
[tex]\displaystyle \mu_s=\frac{F_r}{N}[/tex]
Where Fr is the friction force and N is the normal force exerted by the surface over the object.
In the absence of any other vertical force, the normal is equal to the weight of the object:
[tex]N = W = m.g[/tex]
The box has a mass of m=80 Kg, thus the normal force is:
[tex]N = 80\ Kg * 9.8\ m/s^2[/tex]
N = 784 N
The student needs to push with a force of 400 N to make the box move. That is the force that barely outcomes the friction force. Thus:
[tex]F_r=400\ N[/tex]
Calculating the coefficient:
[tex]\displaystyle \mu_s=\frac{400}{784}[/tex]
[tex]\mathbf{\mu_s=0.51}[/tex]
A ferris wheel with radius 12 m makes a revolution every 3 minutes. Find the linear (tangental) speed of a passenger. How far does a person move in a 5 minute ride?
Answer:
The linear (tangential) speed of a passenger is 0.4188 m/s
The distance traveled by the person in 5 minutes ride is 125.64 m
Explanation:
Given;
radius of the Ferris, r = 12 m
1 revolution per 3 minutes, [tex]\omega = \frac{2\pi (radian)}{3\ (minutes)} *\frac{1\ minute}{60 \ seconds} = 0.0349 \ rad/s[/tex]
The linear (tangential) speed of a passenger is given by;
v = ωr
v = (0.0349)(12)
v = 0.4188 m/s
The distance traveled by the person in 5 minutes ride is given by;
d = vt
d = (0.4188)(5 x 60)
d = 125.64 m