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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The SI unit for weight is ________.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight
Answer:
Newton
Explanation:
The SI unit for weight is newton.
A student in the Biomechanics class has decided that she would like to make her arms
stronger. She has a mass of 63 kg, She chooses to complete some elbow flexion exercises
using a kettlebell. For this problem, consider the hand and forearm to be a single segment.
The distance from her elbow to her wrist is 22.86 cm.
The force from the kettlebell is applied to her hand, which is 30.48 cm from her elbow joint.
She knows that the moment arm of the elbow extensor muscles about the elbow axis is
Answer:
what is heat and transfer
The hot exhaust from a rocket travels in one direction, and the rocket travels in the opposite direction. This is an example of?
A. Equal and opposite forces.
B. Friction.
C. Balanced forces.
D. Inertia.
Answer:
Explanation:
The hot exhaust from a rocket travels in one direction, and the rocket travels in the opposite direction. This is an example of?
A. Equal and opposite forces.
B. Friction.
C. Balanced forces.
D. Inertia.
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 diamond sparkles more than a glass cut to similar shapes,why?
Answer:
because when you shape glass it's not a shiny as diamond
When a penny is dropped, it takes 16 seconds. What is its height
Why do you think we need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies?
We need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies in order to verify the theories and to make laws.
What is theory?A theory is an explanation for observations of the natural world that has been carried by using the scientific method, and which brings together many facts.
So we can conclude that we need evidence to test theories about the composition and origin of our solar system in order to verify the theories and to make laws.
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A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent resistance of the combination of these same two resistors when connected in parallel?
Answer:
R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical
2 R1 = 12 and R1 = R2 = 6 ohms
1 / R = 1 / R1 + 1 / R2 for resistors in parallel
R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3
The equivalent resistance would be 3 ohms if connected in parallel
define what is physics
In my own words, I would say that physics is an area of science that seeks to explain and understand the fundamental nature of the dynamics of objects, essentially defining how objects can interact, in space and in time.
A proton moves north with a speed of 3 x 10^6 m/s. A 5 Tesla magnetic field is directed west. Determine the magnitude and direction of the magnetic field on the proton.
The magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.
Magnitude of magnetic force on the protonThe magnitude of magnitude force on the proton is calculated as follows;
F = qvB sinθ
where;
q is the charge of the protonv is the speed of the protonB is the magnitude of the magnetic filed θ is the angle between the field and speedSubstitute the given parameters and solve for the magnetic force.
F = (1.6 x 10⁻¹⁹) x (3 x 10⁶) x (5) X(sin90)
F = 2.4 x 10⁻¹² N
Thus, the magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.
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A popcorn-maker transfers 250J of energy into other energy stores every second. What is its power?
Now
[tex]\\ \sf\rightarrowtail Power=\dfrac{Work\:done}{Time\:taken}[/tex]
[tex]\\ \sf\rightarrowtail Power=\dfrac{250}{1}[/tex]
[tex]\\ \sf\rightarrowtail Power=250W[/tex]
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Consider the Boeing 787 "Dreamliner", with a mass of 177000 kg. In this particular model, the distance from the front wheels to the rear set of wheels is 21.7 m.
(a) If the center of mass of the airplane is along a line through the center and 3.00 m in front of the rear wheels, how much force, in meganewtons, does the ground exert on each set of rear wheels when the plane is at rest on the runway?
(b) How much force, in meganewtons, does the ground exert on the front set of wheels?
(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.
(b) The force the ground exerts on the front set of wheels is 0.239 MN.
Center mass of the airplaneThe concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.
Some assumptionsThe wheels under the wind do not pass through the center line.The position of the front wheel is constant and it is zero mark (origin).The rear wheels are at 21.7 m markPosition of the center mass of the plane is calculated as follows;
Let the position of the center mass, Xcm = y
the center mass is 3 m in front of rear wheels, that is
21.7 - y = 3
y = 21.7 - 3
y = 18.7 m
Xcm = 18.7 m
Mass of the plane at the position of the rear wheelsLet the mass of the plane at front wheels = M1
Let the mass of the plane at rear wheels = M2
[tex]X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}[/tex]
[tex]18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg[/tex]
Force exerted by the ground on each rear wheelThere are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;
[tex]W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN[/tex]
Mass of the plane at the position of the front wheelM1 + M2 = 177,000
M1 = 177,000 - M2
M1 = 177,000 - 152,529.95
M1 = 24,470.05 kg
Force exerted by the ground on the front wheelW = mg
W = 24,470.05 x 9.8
W = 239,806.5 N = 0.239 MN
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Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
Answer:
Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between
Explanation:
Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.
Hopefully this helps- let me know if you have any questions!
Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
Answer :-Gravitational Force :- It is the attractive force between the two body having some mass.Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this This force helps the celestial bodies to revolve in there orbits Factors effecting Gravitational force:- Mass :- gravitational force directly proportional on the masses of the body between which it is acting [tex]gravitational \: force ∝m{ \tiny1}m{ \tiny2}[/tex]Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses [tex]gravitational \: force∝ \frac{1}{ {r}^{2} } [/tex]Formula of the Gravitational force :-[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]
where
F is gravitational force G is gravitational constant ( universal constant ) m 1 is mass of first body m2 is mass of second body r is the distance between the two bodiesgive the mathematical expression for coulombs force if q1,q2 are the magnitude of charges and r is the distance between them.
Give the mathematical expression for coulomb's force if q1, q2 are the magnitude of charges and r is the distance between them.
F=K q1q2/r2
any help here please ???
Answer:
16 reflection,incidence
A person with a mass of 80 kg is at rest. What is their momentum? Write the
number only. *
Your answer
Answer:
0
Explanation:
because no distance covered by body
How can you describe the relationship between height and pressure?
Answer:
p = rho× g × h
Explanation:
p: pressure
rho : density
g : gravity acceleration
h : height
A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in the xy plane.
What is the electric potential V(z) on the z axis as a function of z , for z>0 ?
What is the magnitude E of the electric field on the z axis, as a function of z , for z>0 ?
The electric potential V(z) on the z-axis is : V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Given data :
V(z) =2kQ / a²(v(a² + z²) ) -z
Determine the electric potential V(z) on the z axis and magnitude of the electric fieldConsidering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex] ----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]
V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]
= [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]
Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
Also
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
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A bucket of mass m is attached to a rope that is wound around the outside of a solid sphere (I = 2/5 M^2) of radius R. When the bucket is allowed to fall from rest, it falls with an acceleration of a down. What is the mass of the sphere in terms of m, R, a, and g?
Answer:
[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)
Explanation:
The forces on the bucket are:
Weight of the bucket: [tex]m\, g[/tex] (downward.)Tension in the rope (upward.)Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:
[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].
The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].
Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:
[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].
At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].
Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].
Therefore, the magnitude of the angular acceleration of this sphere would be:
[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].
The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].
Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].
Hence the equation:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].
Solve this equation for [tex]M[/tex], the mass of this sphere:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].
[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].
[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].
23) What is the magnitude of the electric field intensity at a point where a proton experiences an
electrostatic force of magnitude 2.30 x 10-25 newton?
Answer: I am so sorry i hadn't learned this Yet~!
Explanation:
Fill in the blank with the correct response.
Light striking a mirror at a 45° angle will be reflected at a
A man of mass 70 kg climbs stairs of vertical height 2.5m. Calculate the work done against the force of gravity. (Take g = 9.81 ms?)
Answer:
1716.75 J
Explanation:
Step 1: First check what we are provided with. As per given question we have:
mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².
Step 2: Check what we are asked to find out.
Work done = Change in Potential energy
The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.
Step 3: Substitute the known values in the above formula.
→ P = 70 × 2.5 × 9.81
→ P = 1716.75 J
Hence, the work done against the force of gravity is 1716.75 J.
A student is standing at a distance of 45 m from a wall. He gives a loud clap at the echo is heard 0.3a later. Calculate the speed of sound.
Answer:
300 m/s
Explanation:
2d = vt
v = 2d/t
v = 2×90/.3
v=300 m/s
d = distance
t = time
v = velocity/speed of sound
A spring-loaded ballistic cart measuring 0.68 kg is in contact with a second 0.80 kg
cart. The carts are initially at rest on a level surface. The spring is released and the
lighter cart is observed to move at +0.52 m/s afterward. What is the velocity of the
other cart?
Answer:
wait in comments.................
An experiment is set up as follows:
A mass m = 9.4kg is sent down a frictionless ramp with an initial velocity of 2.5m/s. The ramp is 87cm
long and has an angle of 36°above the horizontal. At the bottom of the ramp is a horizontal surface with a
kinetic friction coefficient of μk = 0.27. At the far side of the horizontal surface, 48cm away, is a spring with
k-constant ks = 3,413.7N
m that will be compressed as the mass collides with the spring. The experiment
ends as the spring is fully compressed and the mass is at rest.
Note: The distance that the spring is compressed is in addition to the 48cm.
7) Find the initial energy of the mass. (10pts)
8) How far will the spring compress if there is no surface friction under the spring? (10pts)
9) How far will the spring compress if the surface friction continues under the spring? (20pts)
(a) The initial energy of the mass is 76.485 J.
(b) The compression of the spring in the absence of friction is 21.2 cm.
(c) The compression of the spring in the presence of friction is 19.4 cm.
Initial energy of the massThe initial energy of the mass is determined as follows;
E = K.E + P.E
E = ¹/₂mv² + mgh
where;
h is the height of the rampE = ¹/₂mv² + mg x Lsinθ
P.E = ¹/₂(9.4)(2.5)² + (9.4)(9.8)(0.87)(sin36)
P.E = 76.485 J
Compression of the spring when there is no surface tensionThe compression of the spring in the absence of friction is calculated as follows;
Ux = E
¹/₂kx² = 76.485
kx² = 2(76.485)
x² = (2 x 76.485)/k
x = √(2 x 76.485)/k
x = √(2 x 76.485 / 3413.7)
x = 0.212 m
x = 21.2 cm
Compression of the spring in presence of frictionThe compression of the spring in the presence of friction is determined by applying the principle of conservation of energy.
E - Fd = Ux
E - μmgd = ¹/₂kx²
76.11 - (0.27 x 9.4 x 9.8 x 0.48) = ¹/₂(3413)x²
64.17 = 1706.5x²
x² = 0.0376
x = √0.0376
x = 0.194
x = 19.4 cm
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The Ice and steam points of a certain thermometer are found to be 20 cm apart. What temperatureis recorded in Celsius when the length of mercury theard is 5cm above the ice point mark?
Which of the following can only be a situation of increasing temperature?
1. The phase change of the material is going from water to solid
2. The height of the object is increasing
3. The objects speed in increasing at a constant rate
4. The average kinetic energy of the particles of a material is increasing
Increasing the temperature causes an increase in the average kinetic energy of the particles of a material.
What is average kinetic energy of particles?The average kinetic energy of particles is the energy possessed by particles due to their constant motion.
The constant motion of particles occurs due to the energy acquired by the particles, when the temperature of the particles increases, the average kinetic energy increases which in turn increases the speed of the particles.
Thus, we can conclude that, increasing the temperature causes an increase in the average kinetic energy of the particles of a material.
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Explain why is dressing table mirror may become dirsty if wiped with a cloth on a warm day.
When you rub a dry cloth across glass, it creates charged static electricity, which attracts little non-charged dust particles.
How can an athlete participating in a 40m sprint modify and improve their performance based on the kinematic variable of speed and acceleration?
The athlete can improve performance by building strength, coordination and balance.
Who is an Athlete?This is an individual who is proficient in sports and other forms of physical exercise.
Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.
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When resting, a person generates about 412005 joules of heat from the body. The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C. If the heat from the person goes only into the water, find the water temperature.
If a person generates about 412005 joules of heat from the body, the water temperature is mathematically given as
t=21.6296C
What is the water temperature.?Question Parameter(s):
The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C
Generally, the equation for the Heat is mathematically given as
Heat gained =Heat loess
Thereofore
mw*cw*(t-2160)=1.5*10^5
[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]
t=21.6296C
In conclusion, the tempreature
t=21.6296C
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