Based in the relationship between static frictional force and coefficient of static friction, the force of static friction is 24 N.
What is friction?Friction is a force that opposes the relative motion of an object moving over another at their surafce of contact.
Frictional force is constant for each type of material. This constant is known as coefficient of friction.
The coefficient of friction is given as follows:
Coefficient of friction = Frictional force/normal reactionFrom the data given:
coefficient of static friction p = 0.60Weight of block = 40 NForce of static friction = 0.60 × 40
Force of static friction = 24 N.
Therefore, the force of static friction is 24 N.
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Water of mass 3 kg at a temperature of 80 ℃ is added to 5 Kg of water at 5 ℃. Calculate the final temperature of the mixture
The final temperature of the mixture is 33.123 °C.
What is temperature?Temperature can be defined as the hotness or coldness of a thing or place.
To calculate the final temperature of the mixture, we use the formula below.
Formula:
Heat gained by the cold water = heat lost by the hot watercm(t₃-t₁) = cm'(t₂-t₃)m((t₃-t₁) = m'(t₂-t₃)......... Equation 1Where:
m = mass of the cold waterm' = mass of the hot watert₁ = Temperature of the cold watert₂ = Temperature of the hot watert₃ = Temperature of the mixture.make t₃ the subject of the equation
t₃ = (mt₁+m't₂)/(m+m')............. Equation 2From the question,
Given:
m = 5 kgm' = 3kgt₁ = 5 °Ct₂ = 80 °CSubstitute these values into equation 2
t₃ = [(5×5)+(3×80)]/(3+5)t₃ = (25+240)/8t₃ = 33.123 °CHence, The final temperature of the mixture is 33.123 °C.
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A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.
7. A 2.0 kg block, starting from rest, is pushed by a
constant force along a frictionless track. The
position of the block as a function of time is
recorded in the data above. The final momentum
of the block is
(A) 0.8 kgm/s
(B) 1.2 kgm/s
(C) 1.6 kgm/s
(D) 3.2 kgm/s
Answer:
(A) 0.8 kgm/s
Explanation:
because of the even ground it would only slow down
Can someone please help me with this assignment, this is due today
Answer:
did you get it done if not lmk I will help you out tomorrow when I get up
Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.
Express your answer in terms of given quantities and ϵ0.
View Available Hint(s)for Part A
F(V)
The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].
Magnitude of the force
The magnitude of the force each plate experiences due to the other plate is determined as follows;
F = U/d
where;
U is potential energy stored in the capacitor[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]
Q = CV
[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]
where;
C is the capacitanceThe capacitance is given as;
[tex]C = \frac{\varepsilon _o A }{d}[/tex]
[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]
Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].
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Determine
i. the total capacitance for the circuit
ii. the total charge stored in the circuit
iii. the charge stored in C9 (3μF)
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
Total capacitance of the circuit
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
C1 and C2 are in series[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]
C1 and C2 are parallel to C3[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]
C(123) is series to C5 and C6[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]
C7 and C8 are in series[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]
Total capaciatnce of the circuitCt + C(78) = 2 μF + 3 μF = 5 μF
Total charge stored in the circuitThe total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
Charge stored in 3μF capacitorQ = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
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As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
Two identical charges are located 1 m apart and feel a 1 N repulsive electric force. What is the charge of each particle.
The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C
AssumptionLet the charge on each particles be q
How to determine the charge Final force (F) = 1 NDistance apart (r) = 1 mElectrical constant (K) = 9×10⁹ Nm²/C²Charge on 1st particle (q₁) = q =? Charge on 2nd particle (q₂) = q =?The charge on each particle can be obtained by using the Coulomb's law equation as shown below:
F = Kq₁q₂ / r²
F = Kq² / r²
1 = (9×10⁹ × q²) / 1²
1 = 9×10⁹ × q²
Divide both side by 9×10⁹
q² = 1 / 9×10⁹
Take the square root of both side
q = √(1 / 9×10⁹)
q = 1.05×10¯⁵ C
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The ________ of an object is the same on the earth as it is on the moon.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight
Answer:
mass
Explanation:
The mass of an object is the same on the earth as it is on the moon.
Mass always remains constant, it never changes with respect to place.
a=5i+4j-6k ,b=-2i+2j+3k ,c=4i+3j+2k. find the vector perpendicular to a and c
Answer:
Explanation:
You can use the cross product. Let the vector that perpendicular to a and c is [tex]\vec{d}[/tex], so:
[tex]\vec{d}=\vec{a}\times\vec{c}=\left|\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\5&4&-6\\4&3&2\end{array}\right] \right|=(8+18)\hat{i}-\hat{j}(10+24)+\hat{k}(15-16)=26\hat{i}-34\hat{j}-\hat{k}[/tex]
To check that c is perpendicular with a and b, do the dot product between c and a and also c and b and if the result is zero, you're true.
[tex]\vec{d}.\vec{a}=(26*5)-(34*4)+(6)=0[/tex] (c perpendicular to a)
[tex]\vec{d}.\vec{c}=(4*26)-(34*3)-(2*1)=0[/tex] (d perpendicular to c)
Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?
There is no temperature change which drives heat flow, thus no heat will be released by the water.
Heat released by the water when it freezesThe heat released by the water when it freezes is calculated as follows;
Q = mcΔФ
where;
m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - ФiInitial temperature of water, Фi = 0 °C
when water freezes, the final temperature, Фf = 0 °C
Q = 22 x 4200 x (0 - 0)
Q = 0
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
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a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning
Answer:
Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.
Explanation:
A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap
Answer:
A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap
Q. 1 MWH is equal to ------- joules
a.3.6*10^10
b.3.6*10^6
c.3.6*10^9
d.3.6
A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA hangs vertically from the pulley. The mass is released from rest. show answer Incorrect Answer 25% Part (a) What is the magnitude of the tangential acceleration of the hanging mass?
The magnitude of the tangential acceleration of the hanging mass is 2mg/MR
Tangential acceleration of the hanging massThe tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;
τ = Iα
Where;
I is the moment of inertiaα is the angular velocity[tex]\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}[/tex]
Where;
m is the hanging massM is the mass of solid diskThe tangential acceleration is calculated as follows;
[tex]a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}[/tex]
Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR
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what is one type of compact star with a mass similar to the sun but a diameter similar to earth?
Explanation:
neutron star, any of a class of extremely dense, compact stars thought to be composed primarily of neutrons. Neutron stars are typically about 20 km (12 miles) in diameter. Their masses range between 1.18 and 1.97 times that of the Sun, but most are 1.35 times that of the Sun.
A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed 4.20 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 8.40 block recoils with a speed of 0.400 m/s. In the figure; the blocks are in contact for 0.200 s.
For A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed of 4.20 m/s is mathematically given as
F = 193.2N
What is the magnitude of the average force on the 8.40-kg block, while the two blocks are in contact, is closest to?
Generally, the equation for the magnitude of the average force mathematically given as
F = m(v1+v2)/t
F = 8.40(4.2+O.4)/t
F = 193.2N
In conclusion magnitude of the average force is
F = 193.2N
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17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus
Answer: b or d
Explanation: b or d
Answer:
a
Explanation:
F= ma
interestingly
when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases
(man, PHYSICS IS JUST THE BESY)
A: a
object with the smallest mass has largest acceleration
Find the temperature
Answer:
-------1
Explanation:
beacuse that is what i know
. Radiation travels at the speed of light
T or F?
Answer:
electromagnetic radiation moves at the speed of light
An electron has a mass of 9.1x10^-31 kg and a charge
of -1.6x10^-19 C. Suppose you could isolate one electron in
a perfect vacuum and then create an electric field to pull
and edi 107. upward on the electron. How strong would the field have to
be to counteract the electron's weight? (In other words,
di Delfine bu how strong would the field have to be to put the electron in
a state of force equilibrium?)
is one of these the question
1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,
a. Sketch the signal for -4
b. Find the fundamental period of the signal (if it is periodic).
c. Determine if the signal is odd / even or neither.
d. Compute the energy of the signal for all time.
e. Compute the power of the signal for all time.
Given that the function of the wave is f(x) = cos(π•t/2), we have;
a. The graph of the function is attached
b. 4 units of time
c. Even
d. 4.935 J/kg
e. 1.234 W/kg
How can the factors of the wave be found?a. Please find attached the graph of the signal created with GeoGebra
b. The period of the signal, T = 2•π/(π/2) = 4
c. The signal is even, given that it is symmetrical about the y-axis
d. The energy of the signal is given by the formula;
[tex] \frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda[/tex]
Which gives;
E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg
e. The power of the wave is given by the formula;
E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg
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Draw free body diagrams for the following objects: (12pts)
1A) A coaster sitting under a cup of coffee.
1B) A car slowing down as it approaches a stop sign.
1C) Your test stuck to your fridge by a magnet.
1D) A baseball just before it leaves the bat.
(a) The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.
(b) The force diagram of a car slowing down as it approaches a stop sign includes force of the car and frictional force opposing the motion.
(c) The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.
(d) The force diagram of baseball before it leaves the bat incudes only the weight of the baseball acting downwards.
Force diagram of coaster sitting under a cup of coffeeThe force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.
↑ Fn
Ф Fn = W
↓ W
Where;
W is weight of the coaster plus weight of coffeFn is the normal reactionForce diagram of a car slowing down as it approaches a stop signThe force diagram inlcudes the applied force and frictional force opposing the motion.
Ff ← Ф → F
where;
Ff is the kinetic frictional forceF is force of the carForce diagram of test stuck to your fridgeThe force diagram includes the force of the test and action of the fridge which are eqaul and opposite.
Fb ← Ф → Fa
where;
Fa is the force of the testFb is the force of the fridgeForce diagram of baseball before it leaves the batThe force diagram includes only the weight of the baseball acting downwards.
Ф
↓
W = mg
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A 4500 kg Aston Martin traveling at 102 m/s has to stop short because some ducklings
hazard onto the road. The Aston Martin was able to stop in 1.77 seconds. How much
force was placed on the car?
Answer:
-259322.03N
Explanation:
[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]
2. A tennis ball machine launches balls horizontally with an initial speed of 5.3 m/s, from a height of 1.2 m.
a) What will the time of flight be for a tennis ball launched by the ball machine? (3)
b) What will the range of the tennis ball be? (2)
c) What will be the final velocity of the ball with which it reaches the ground? (3)
(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.
(b) The range of the tennis ball be is 1.01 m.
(c) The final velocity of the ball with which it reaches the ground is 7.16 m/s.
Time of flight of tennis ballThe time of flight of the tennis ball is calculated as follows;
h = vt + ¹/₂gt²
1.2 = 5.3t + 0.5(9.8)t²
1.2 = 5.3t + 4.9t²
4.9t² + 5.3t - 1.2 = 0
a = 4.9, b = 5.3, c = 1.2
solve using quadratic formula
t = 0.19 s
Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.
Range of the tennis ballThe range of the tennis ball is calculated as follows;
R = vt
R = 5.3 x 0.19
R = 1.01 m
Final velocity of the ballThe final velocity of the ball with which it reaches the ground is calculated as follows;
vf = vo + gt
vf = 5.3 + 9.8(0.19)
vf = 7.16 m/s
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A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per meter of length. By constructing a suitable Gaussian surface around the wire, Find the magnitude and direction of the electric field at points
a)5.5m
b)2.5m
perpendicular from the center of the wire.
(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.
(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.
Electric field at a point on the Gaussian surface
The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;
E = λ/2πε₀r
where;
λ is linear charge densityε₀ is permitivity of free spacer is the position of the chargeAt a distance of 5.5 m[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C[/tex]
At a distance of 2.5 m[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C[/tex]
Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.
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list out the use of simple machine
Explanation:
simple machine can multiplayer of speed and force
A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after being lifted 2.00 m ? Solve this problem using work and energy
The final speed of the block, after being lifted 2.00 m is 3.39 m/s
What is speed?Speed can be defined as the rate of change in the distance of a body.
To calculate the speed of the block after being lifted 2.00 m, first, we need to calculate the acceleration of the block using the formula below
Formula:
T-mg = ma......... Equation 1Where:
T = Tension in the cablem = mass of the cablea = accelerationg = acceleration due to gravityRestructuring the formula above,
a = (T-mg)/m............... Equation 2From the question,
Given:
T = 19 Nm = 1.5 kgg = 9.8 m/s²Substitute these values into equation 2
a = [(19)-(1.5×9.8)]/1.5a = 4.3/1.5a = 2.87 m/s²Finally, to calculate the speed of the block, we use the formula below.
v² = u²+2as.......... Equation 3Where:
v = Final speedu = initial speeda = accelerations = distanceFrom the question,
Given:
u = 0 m/sa = 2.87 m/s²s = 2.00 mSubstitute these values into equation 3
v² = 0²+(2×2×2.87)v² = 11.48v = √11.48v = 3.39 m/sHence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.
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Mention the objective of the Experiment?
Answer:
I don't understand your question ❓,the object.....of what experiment
Terry is walking down the street at 3 m/s. If he
has a mass of 70 kg, what is his momentum?
[tex]\text{Given that, mass m = 70 kg and velocity v = 3 m/s}\\\\\text{Momentum,}~ p = mv = 70 \times 3 = 210~ kg ~ms^{-1}[/tex]