Answer:
1. Friction enables us to walk freely.
2. It helps to support ladder against wall.
3. It becomes possible to transfer one form of energy to another.
4. Objects can be piled up without slipping.
Explain the meaning of work done and give examples
Answer:
the situation in which the force or any kinds of things rhat are done by force is called work done.ex.typed,wrote
Answer:
WORK DONE on a body is the product ofFORCE IN THE DIRECTION OF DISPLACEMENT AMD DISPLACEMENT.WORK ONLY HAS MAGNITUDE AND NO DIRECTION.
FORCE applied IN the direction of displacement × displacement = POSITIVE WORK DONEFORCE applied OPPOSITE to the direction of displacement × displacement = NEGATIVE WORK DONEFORCE applied PERPENDICULAR to the direction of displacement × displacement = ZERO WORK DONEhope it helps
have a nice day
Instantaneous speed is...
a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.
PLEASE HURRY
Answer:
The speed attained at a particular instant in time.
Explanation:
Instantaneous speed is the speed attained at a particular instant in time.
It is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
It is equal to the rate of change of speed.
It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.
Hence, the correct option is (b).
a car runs of a road and collides with a tree. glass pieces from the windscreen are projected forward and are found an average distance of 12m from the car. the average height of the windscreen is 1.2m.
Establish the speed of the car at the time of impact. assume g=10 m/s²
Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
The geometric shape of the molecule H2,O. A. bent B. Trigonal planar C. linear D. t shape
Answer:
A. bent
Explanation:
Water molecule (H2O) is said to contain four valence electron pairs (2 bond pairs and 2 lone pairs). The presence of lone pair of electrons in the water molecule influences its molecular geometry or shape.
Since water has two lone pairs of electrons, which repel each other according to the VSEPR theory, water molecule is said to have a BENT molecular geometry.
help plzzzzzzzzzzzz ?
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?
Answer:
v₀ = 2.67 m / s
Explanation:
This problem can be solved using the Kinetic Enemy Work Theorem
W = ΔK
Work is defined by the relation
W = fr. d
The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.
the friction force opposes the displacement therefore its angle is 180º
W = - fr d
Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane
Y axis
N- Wy = 0
N - W cos tea = 0
the friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
W = - μ W cos θ d
let's look for kinetic energy
the minimum velocity at the highest point is zero
K_f = 0
the initial kinetic energy is
K₀ = ½ m v₀²
we substitute energy in the work relationship
- μ W cos θ d = 0 - ½ m v₀²
v₀² = - μ W cos θ 2d / m
Let's use trigonometry to find distance d
sin θ= y / d
d = y /sin θ
d = 3.50 / sin 30
d = 7 m
let's calculate
v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50
v₀ = √7.129
v₀ = 2.67 m / s
What is the magnitude of the resultant velocity for a bird flying first at a
speed of 10 m/s North East and then flying to South at a speed of 8 m/s?
Answer:
the magnitude of resultant is landify space ociured by meteroid
The magnitude of resultant velocity for a bird flying is 7.132 m/s.
What is Velocity?Velocity is defined as the directional motion of an object which is observed from a particular frame of reference and measured by a particular standard of time, as an indication of the rate of change of position.
Velocity is the Vector quantity as it has magnitude as well as direction. It can be represented as
[tex]\overline{v}={\frac{\Delta x}{\Delta t}}[/tex]
[tex]\overline{v} = average velocity[/tex]
[tex]{\Delta x} = displacement[/tex]
[tex]{\Delta t} = change in time[/tex]
For the information above, two vectors are represented as the sides of a parallelogram, then the resulting vector is given as the diagonal of that parallelogram. The resultant vector is that it can have any value between a maximum and a minimum depending on the angle; But scalar numbers can have either maximum or minimum value depending on whether they are added or subtracted.
The resultant vector of any two vectors is given by: [tex]v^2= \sqrt{v_1^2+ v_2^2+ 2 v_1v_2 cos }[/tex]Θ
where, Θ is the angle between these two vectors.
The angle between directions north-east and south is 135°.
So, the magnitude of the resultant vector is
[tex]v= \sqrt{10^2 + 8^2 +2.10.8. cos 135}[/tex]
v= 7.132 m/s
Thus, the magnitude of resultant velocity for a bird flying is 7.132 m/s.
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They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?
Answer:
The efficiency of Carnot's heat engine is 26.8 %.
Explanation:
Temperature of hot reservoir, TH = 100 degree C = 373 K
temperature of cold reservoir, Tc = 0 degree C = 273 K
The efficiency of Carnot's heat engine is
[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]
The efficiency of Carnot's heat engine is 26.8 %.
A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
a) how fast is book traveling the instant before it impacts the ground below the cliff?
b) how far above the bottom of the cliff is the object moving at 12 m/s?
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
[tex]v^2=v_0^2+2a[/tex]Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
[tex]v^2=0^2+2(-9.8)(-120)[/tex] so
[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
[tex]12^2=0^2+2(-9.8)[/tex]Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.
A boy observes a fish at a depth of 4 metres under the surface of a lake. If the refractive index of the water is 4/3. what is the apparent depth of the fish as seen by the boy? Apparent depth Real depth
Answer:
[tex]{ \tt{n _{w} = \frac{real \: depth}{aparent \: depth} }} \\ \\ \frac{4}{3} = \frac{4}{d} \\ { \tt{d = 3 \: meters}}[/tex]
A boy observes a fish at the depth of 4 meters under the water's surface. The refractive index of the water is 4/3 then the apparent depth of the fish as seen by the boy is 3 meters.
What is the Refractive index?Refractive index is also known as the refraction, is a measure of the way a light beam bends when it goes through different media. The refractive index n is computed by dividing the sine of the angle of incidence by the sine of the angle of refraction, i.e., n = sin i/sin r, if I will be the angle of incidence of a light source in a void (angle between the inbound ray and the normal, or perpendicular to the surface of a medium).
The Index of refraction is also equal to c/v, or the ratio of a wavelength of light's velocity in a substance to its velocity in a void.
As per the given data in the question,
n(w) = real depth/apparent depth
4/3 = 4/d
d = 3 meters.
Therefore, the apparent depth will be equal to 3 meters.
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You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed end, and a 312 Hz tuning fork. The piece of pipe resonates in the 2nd resonant length when it is adjusted to a length of 81.0 cm. What is the speed of sound on that day?
Answer:
336.96m/s
Explanation:
answer is in photo above
The speed of sound on that day is 336.96 m/s.
What is speed?Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.
given that:
frequency of the sound = 312 Hz.
2nd resonant length = 81.0 cm.
If the wavelength of sound is λ; the 2nd resonant length of plastic pipe with an adjustable closed end = 3λ/4
Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.
So, the speed of sound on that day is = frequency × wavelength
= 312 Hz × 0.81 ×(4/3) m.
= 336.96 m/s.
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An airplane is traveling at 940 km/h. How long does it take to travel 2.00 km?
Answer:
Explanation:
Since the path an airplane travels is pretty much horizontal, then acceleration in this dimension is 0 and the equation we use is strictly the d = rt one. Here we use it and solve for t:
[tex]t=\frac{d}{r}[/tex] and filling in:
[tex]t=\frac{2.00}{940}[/tex] so
t = .002 hrs which is the same as 7.2 seconds
A student wishes to construct a mass-spring system that will oscillate with the same frequency as a swinging pendulum with a period of 3.45 S. The student has a spring with a spring constant of 72.0 N/m. What mass should the student use to construct the mass-spring system?
Answer:
21.73 kg
Explanation:
Applying,
T = 2π√(m/k)............... Equation 1
Where T = period, m = mass on the spring, k = spring constant, π = pie.
make m the subject of the equation
m = T²k/4π²................. Equation 2
From the question,
Given: T = 3.45 s, k = 72.0 N/m, π = 3.14
Substitute these values into equation 2
m = (3.45²×72)/(4×3.14²)
m = 21.73 kg.
Hence the mass should be 21.73 kg
Why do some "people" still believe that the earth is "flat".
I called my brother in Australia and it was day here, and night there, HOW CAN THE EARTH POSSIBLY BE FLAT, i did a zo,om meeting to prove that it was live. WHAT ARE PE,OPLE TH,INKING!!!
Answer: Members of the Flat Earth Society claim to believe the Earth is flat. Walking around on the planet's surface, it looks and feels flat, so they deem all evidence to the contrary, such as satellite photos of Earth as a sphere, to be fabrications of a "round Earth conspiracy" orchestrated by NASA and other government agencies.
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Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite ends of the same rope and pull the other toward him. The magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration. What is the ratio of Hank's mass to Harry's mass?
Answer:
the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1
Explanation:
Given the data in the question;
Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.
We know that from Newton's Second Law;
Force = mass × Acceleration
F = ma
Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.
Now,
Mass[tex]_{Hank[/tex] × Acceleration[tex]_{Hank[/tex] = Mass[tex]_{Henry[/tex] × Acceleration[tex]_{Henry[/tex]
so
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = Acceleration[tex]_{Henry[/tex] / Acceleration[tex]_{Hank[/tex]
given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = 1 / 1.26
Mass[tex]_{Hank[/tex] / Mass[tex]_{Henry[/tex] = 0.7937 or [ 0.7937 : 1 ]
Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]
Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))
Explanation:
it depens on the subject and object. Let's example
you are driving a tesla car with your dog sitting on your side. You will say that the dog is at REST
but your friend, standing in sidewalk, seeing the same dog, will say that your dog is moving because it has MOTION from your car
Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ball’s journey will gravity be the greatest force acting on the ball?
Answer:
If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.
40 ohms
1.2 A
40 ohms
12 V
Calculate the total energy developed in 5
minutes by the system above.
Answer:
17280 J and 1080 J
Explanation:
Given :
R= 40 ohm
I=1.2A
t= 5 min=60×5=300 sec
Now,
Total energy can be calculated as:
[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]
Now,
V=12V
R=40 Ohm
[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]
Total energy is 17280 J and 1080 J
What actually heats up the atmosphere?
Answer:
The heat source for our planet is the sun. Energy from the sun is transferred through space and through the earth's atmosphere to the earth's surface. Since this energy warms the earth's surface and atmosphere, some of it is or becomes heat energy.
What is the resistance of a bulb of 4ow
connected in a line of 220v?
2
Answer:
1210 ohm
Explanation:
Given :
P=40 W
V=220 V
Now,
[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]
Therefore, resistance of bulb will be 1210 ohm
A clay vase on a potter's wheel experiences an angular acceleration of 7.24 rad/s2 due to the application of a 13.3-N m net torque. Find the total moment of inertia of the vase and potter's wheel.
Answer:
[tex]I=1.83\ kg-m^2[/tex]
Explanation:
Given that,
The angular acceleration of the wheel,[tex]\alpha =7.24\ rad/s^2[/tex]
Net torque,[tex]\tau=13.3\ N-m[/tex]
We need to find the total moment of inertia of the vase and potter's wheel. We know that,
Net torque,[tex]\tau=I\alpha[/tex]
Where
I is the moment of inertia
So,
[tex]I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{13.3}{7.24}\\\\I=1.83\ kg-m^2[/tex]
So, the moment of inertia of the vase is equal to [tex]1.83\ kg-m^2[/tex].
A given wave has a wavelength of 1.4 m and a frequency of 2.0 Hz. How fast
is the wave moving?
O A. 0.6 m/s
O B. 3.4 m/s
O C. 0.7 m/s
D. 2.8 m/s
Answer:
The wave speed is 2.8 m.
Explanation:
Wavelength = 1.4 m
frequency, f = 2 Hz
the wave speed is given by
wave speed = wavelength x frequency
wave speed = 1.4 x 2 = 2.8 m
option (D) is correct.
Which of the following is a good example of a contact force?
ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up
Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
What is the answer can you explain it to me
Answer:
C) 300 Ohm.
Explanation:
In a series circuit, total resistance is just adding all the resistance together. So R (total) = 75 +75 +75+ 75 = 300 ohms
Parallel circuit are different because you add the inverses of resistance and you flip the final answer.
You can confirm your answers using the tools below:
https://www.omnicalculator.com/physics/series-resistor
https://www.omnicalculator.com/physics/parallel-resistor
Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v of the car, assuming that the tires are not slipping against the ground.
Answer:
the linear speed of the car is 28.83 m/s
Explanation:
Given;
radius of the car, r = 0.33 m
angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s
The linear speed of the car is calculated as;
V = ωr
V = 87.35 rad/s x 0.33 m
V = 28.83 m/s
Therefore, the linear speed of the car is 28.83 m/s
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal speed be right before it hits the ground?
A. 15 m/s
B. 0 m/s
C. 30 m/s
D. 40 m/s
Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s
A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sled attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Answer:
[tex]\mu=0.185[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=17kg[/tex]
Force [tex]F=33N[/tex]
Velocity [tex]v=1.6m/s[/tex]
Distance [tex]d= 9.8m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=\triangle K.E+\triangle P.E[/tex]
Where
[tex]\triangle K.E=(F-F_f)*2[/tex]
[tex]F_f=F+\frac{\triangle K.E}{d}[/tex]
[tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]
[tex]F_f=30.8N[/tex]
Since
[tex]f = \mu*m*g[/tex]
[tex]\mu= 30.8/(m*g)[/tex]
[tex]\mu= 30.8/(17*9.81)[/tex]
[tex]\mu=0.185[/tex]
How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
An Na ion has a charge of +1 and an Oion has a charge of -2. How many Na ions does one Oion need to balance
charges?