Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.
An organism has 20 chromosomes after fertilization.after meiosis, how many chromosomes would each sex cell have?
Answer:
EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n
means haploid parent cells join or fuse to form diploid zygote
Answer:
10
Explanation:
An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?
Answer:
Ff = 33.4N
Explanation:
To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.
The frictional force is given by:
[tex]F_f=\mu_kN[/tex] (1)
Ff: frictional force = ?
µk: coefficient of kinetic friction = 0.167
N: normal force of the object = 200N
You replace the values of the parameters in the equation (1):
[tex]F_f=(0.167)(200N)=33.4N[/tex]
The frictional force, while the objects is moving, is 33.4N
The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?
Answer:
17.85°
Explanation:
To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
n1: index of refraction of Sophia's cornea = 1.387
n2: index of refraction of aqueous fluid = 1.36
θ1: angle to normal in the first medium = 17.5°
θ2: angle to normal in the second medium
You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:
[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]
hence, the angle to normal in the aqueous medium is 17.85°
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
[tex]^{\circ}C=K-273[/tex]
Put T = 100 K
[tex]T=100-273=-173^{\circ} C[/tex]
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
Can someone help me with this question
Answer:
hypothesis , hope it helps
Explanation:
Answer:
Inference
Explanation:
Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.
Gas is contained in a piston-cylinder assembly and undergoes three processes. First, the gas is compressed at a constant pressure of 100 [kPa] from initial volume of 1.0 [m3] to a volume of 0.5 [m3]. Second, the gas pressure is increased by heating at constant volume up to 200 [kPa]. Third, the gas is returned to its initial pressure and volume by a process for which P ∀=constant. All pressures given are absolute. For the gas as a system, is the system best considered open, closed, or isolated? Why?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is B
Explanation:
The system is best considered a closed system because looking at process we can see that there was no exchange of matter between the system and the surrounding,(as the was no escape of matter from the system to the surrounding )
Secondly we can deduce that there is a variation in the volume. from [tex]1.0 m^3[/tex] to [tex]0.5 m^3[/tex]
At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.
Answer:
1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Explanation:
The kinetic friction works against the kinetic energy of the car and the car stops when these two equalises .
friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.
= μk x mg
work done by friction
= force x displacement
= μk x mg x d
kinetic energy of car at the time of accident = 1/2 m v²
kinetic energy = work done by friction
1/2 m v² = μk x mg x d
d = v² / (2 μk x g)
v² = 2dμk g
v = √(2dμk g)
Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sled attains a speed of 2.0 m/s in 8.5 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. Num
Answer:
Coefficient of kinetic friction = 0.146
Explanation:
Given:
Mass of sled (m) = 18 kg
Horizontal force (F) = 30 N
FInal speed (v) = 2 m/s
Distance (s) = 8.5 m
Find:
Coefficient of kinetic friction.
Computation:
Initial speed (u) = 0 m/s
v² - u² = 2as
2(8.5)a = 2² - 0²
a = 0.2352 m/s²
Nweton's law of :
F (net) = ma
30N - μf = 18 (0.2352)
30 - 4.2336 = μ(mg)
25.7664 = μ(18)(9.8)
μ = 0.146
Coefficient of kinetic friction = 0.146
Convert from scientific notation to standard form
9.512 x 10-8
Answer:
0.00000009512
Explanation:
Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.
let's pass the number 9,512 10⁻⁸ to decimal notation
9,512 / 10⁸ = 9,512 / 100000000
0.00000009512
As we see writing this number, it is very easy to make mistakes
A soccer player is benched for being late to the game. In a fit of anger, she drops her ball from the top of the Physics building. It falls 4.9 meters after 1.0 second has elapsed. How much farther does it fall in the next 2.0 seconds
Answer:
The distance is [tex]S = 39.2 \ m[/tex]
Explanation:
From the question we are told that
The distance covered after t = 1 s is [tex]d = 4.9 \ m[/tex]
According to the equation of motion
[tex]v^2 = u^2 + 2ad[/tex]
Now u = 0 m/s since before the drop the ball was at rest
[tex]v^2 = 2ad[/tex]
here [tex]a =g = 9.8 \ m/s^2[/tex]
So
[tex]v = 9.8 m/s[/tex]
Also from equation of motion we have that
[tex]S = ut + \frac{1}{2} at^2[/tex]
Now at t = 2 s , as given from the question
Then u = v = 9.8 m/s
And
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 39.2 \ m[/tex]
You are on a train traveling east at speed of 19 m/s with respect to the ground. 1)If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground
Answer:
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
Explanation:
Relative velocity with respect to the ground is;
Vbg = velocity of train with respect to the ground + your velocity with respect to the train
Vbg = Vtg + Vbt ......1
Given;
velocity of train with respect to the ground;
Vtg = 19 m/s
your velocity with respect to the train;
Vbt = 1.5 m/s
Substituting the given values into the equation 1;
Vbg = 19 m/s + 1.5 m/s
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
A 2500 kg truck moving at 10.00 m/s strikes a car waiting at the light. Assume there is no friction on the road. The hook bumpers continue to move at 7.00 m/s. What is the mass of the struck car
URGENT : Which of the following is the most stable isotope? Explain.
Answer:
Plutonium–238
Explanation:
The stability of isotopes is largely dependent on their half-life.
Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.
From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.
We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.
Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.
70 pointss yall !!! helpp
Answer:
A= The type of plant
B= How tall the plant is
Explanation:
Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.
Penny is adjusting the position of a stand up piano of mass mp = 150 kg in her living room. The piano is lp = 1.35 m in length. The piano is currently at an angle of θp = 36 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano.
Required:
a. Write an expression for the minimum magnitude of the force FS in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is µs.
b. The coefficient of kinetic friction between the carpet and the piano is uk = 0.27. Once the piano starts moving, calculate the torque τ in N·m that Penny needs to apply to keep moving the piano at a constant angular velocity.
c. Calculate the amount of work Wp, in J Penny does on the piano as she rotates it.
Answer:
a) Fs = (μs*mp*g)/2 | b) τ = Fs*lp | c) Wτ,constant = τΘ
Explanation:
a) Fs = (μs*mp*g)/2
b) τ = Fs*lp
c) Wτ,constant = τΘ
A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of the tooth on the prey, in N/cm², is
a) 0.0013 N/cm²
b) 128 N/cm²
c) 320 N/cm²
d) 640 N/cm²
Answer:
640N/cm^2Answer D is correct
Explanation:
[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]
hope this helps
brainliest appreciated
good luck! have a nice day!
Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?
Answer:
a) v₀ = - 164.62 m / s , c) y = 122.5 m
Explanation:
We can solve this exercise using the free fall kinematic relations.
We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m
For the boy
y = y₀ + v₀ t - ½ g t²
with free fall its initial speed is zero
y = ½ g t2
For superman
y = y₀ + v₀ (t-5) - ½ g (t-5)²
how superman grabs the lot just before hitting the ground
we look for the time it takes the boy down
t = √ (2 y₀ / g)
t = √ (2 180 / 9,8)
t = 6.06 s
in the equation for superman, we clear the volume and calculate
v₀ (t-5) = -y₀ + ½ g (t-5)²
v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²
v₀ 1.06 = -174.49
v₀ = - 174.49 / 1.06
v₀ = - 164.62 m / s
the negative sign indicates that the initial speed is down
b) to graph the position of the two we use the table
t (s) Y_boy (m) Y_superman (m)
0 180 180
1 175.1 180
5 57.5 180
6 3.6 10.18
see attachment for the two curves
c) calculate the height that falls a lot in the 5 seconds (t = 5)
y = -1/2 g t²
y = ½ 9.8 5²
y = 122.5 m
for this height superman has not yet left the skyscraper, so the boy hits the ground
A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller, or equal to that observed inside the bus.
e. is greator than that observed inside the bus
Answer:
d
Explanation:
good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed
assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.
so a is correct
now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe
b is correct
assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus
e also correct
so correct answer is d. it depends on the speed of ball tossed by the student in front.
In the Life Cycle of Stars diagram, what stage does letter J represent?
A.) white dwarf
B.) black dwarf
C.) black hole
D.) neutron star
Which letters in the Life Cycle of Stars diagram represent stars on the main sequence?
A.) F & I
B.) C & G
C.) A & E
D.) B & D
In the Life Cycle of Stars diagram, what stage does letter L represent?
A.) neutron star
B.) black hole
C.) white dwarf
D.) black dwarf
In the Life Cycle of Stars diagram, what stage does letter I represent?
A.) neutron star
B.) black dwarf
C.) black hole
D.) white dwarf
In the Life Cycle of Stars diagram, what does letter D represent?
A.) a high mass star
B.) a white dwarf
C.) a cool star
D.) a low mass star
In the Life Cycle of Stars diagram, what stage does letter C represent?
A.) nuclear fusion
B.) a supernova
C.) a planetary nebula
D.) protostar formation
Which letter in the Life Cycle of Stars diagram represents a star forming region of space?
A.) M
B.) H
C.) J
D.) G
Which letter in the Life Cycle of Stars diagram represents a planetary nebula?
Group of answer choices
A.) G
B.) H
C.) L
D.) M
ANSWER: num 1 is black hole
why is India called peninsula?
Answer:
India is a peninsula.
Explanation:
India is called as Indian Peninsula because it is surrounded by the Indian ocean on the south, the Arabian sea on the west and the Bay of Bengal on the east.
g 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
Answer:
Angle = 18.41°
Explanation:
Torque = F•r•sin θ
where;
F = force
r = distance from the rotation point
θ = the angle between the force and the radius vector.
We are given;
Torque = 15 N.m
F = 95 N
r = 0.5 m
Thus, plugging in the relevant values ;
15 = 95 × 0.5 × sin θ
sin θ = 15/(95 × 0.5)
sin θ = 0.3158
θ = sin^(-1)0.3158
θ = 18.41°
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.
Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?
Answer:
The lifetime of the particle is [tex]\Delta t = 2.6*10^{-25} \ s[/tex]
Explanation:
From the question we are told that
The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]
The intrinsic width is [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]
The lifetime is mathematically represented as
[tex]\Delta t = \frac{h}{\Delta E}[/tex]
Where h is the Planck's constant with a value of [tex]1.055*10^{-34} \ J\cdot s[/tex]
substituting values
[tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]
[tex]\Delta t = 2.6*10^{-25} \ s[/tex]
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2. What is the magnitude of the force exerted on the middle cube by the left cube in this case
Answer:
24 Newtons
Explanation:
The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.
The mass of those two cubes combined is 6 + 6 = 12 kg
So, using the following equation, we can find the force:
Force = mass * acceleration
Force = 12 * 2
Force = 24 Newtons
A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of the string is 0.62 m, and the tension in the string when the ball is at the top of the circle is 4.0 N. What is v
Answer:
Explanation:
Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force
T + mg = m v² / r
4 + .25 x 9.8 = .25 x v² / .62
6.45 = .25 v² / .62
v² = 16
v = 4 m /s .
A corpse is discovered in a room that has its temperature held steady at 25oC. The CSI ocers ar- rive at 2pm and the temperature of the body is 33oC. at 3pm the body's temperature is 31oC. Assuming Newton's law of cooling and that the temperature of the living person was 37oC, what was the approximate time of death
Answer: Around 0:35 Pm or 12:35 Am
Explanation:
The equation that describes the cooling of objects can be written as:
T(t) = Ta + (Ti - Ta)*e^(k*t)
Where Ta is the ambient temperature, here Ta = 25°C.
Ti is the initial temperature of the body, we have Ti = 37°C.
t is the time.
k is a constant.
So our equation is:
T(t) = 25°C +12°C*e^(k*t)
at 2pm, the temperature was 33°C
at 3pm, the temperature was 31°C.
we want to find the hour where we have our t = 0, suppose this hour is X.
then we can write our times as:
2pm ---> 2 - X
3pm ----> 3 - X
and our equations are:
33°C = 25°C + 12°C*e^(k2 - k*X)
31° = 25°C + 12°C*e^(k3 - k*X)
So we have two equations and two variables, let's solve the system.
first, simplify it a bit, for the first eq:
33 - 25 = 12*e^(k2 - k*X)
8/12 = e^(k2 - k*X)
ln(8/12) = k*2 - k*X
for the second equation we have:
31 - 25 = 12*e^(k3 - k*X)
6/12 = e^(k3 - k*X)
ln(6/12) = k*3 - k*X
So our equations are:
1) ln(2/3) = 2*k - X*k
2) ln(1/2) = 3*k - X*k
First, let's isolate one of the variables in one of the equations. let's isolate k in the first equation.
ln(2/3)/(2-X) = k
now we can replace it in the second equation:
ln(1/2) = 3*ln(2/3)/(2 - X) - X*ln(2/3)/(2-X)
now let's solve it for X, i will take a = ln(1/2) and b = ln(2/3) so it is easier to read.
a = 3*b/(2 - X) - X*b/(2 - X)
a*(2 - X) = 3*b - X*b
2a - aX = 3b - Xb
X(a - b) = 2a - 3b
X = (2*ln(1/2) - 3*ln(2/3))/(ln(1/2) - ln(2/3)) = 0.590
now, knowing that one hour has 60 minutes, then this is:
0.59*60m = 35 minutes
So the hour of death is 0:35 Pm or 12:35 Am
A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.30 rev/s in 3.05 s . Part A What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s
Answer:
[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
Explanation:
Angular acceleration
[tex]\begin{aligned}
\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\
\omega_{i} &=0 \\
\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\
&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&=27.02 \mathrm{rad} / \mathrm{s} \\
\alpha &=\frac{(27.02-0)}{3.15} \\
&=8.57 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
a)Tangential acceleration
[tex]\begin{aligned}
a &=r \alpha \\
&=\frac{12}{2} \times 10^{-2} \times 8.57 \\
a &=0.51 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
This question involves the concepts of the equations of motion for angular motion.
The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".
First, we will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
t = time taken = 3.05 s
Therefore,
[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]
Now, the tangential acceleration can be given as follows:
[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]
a = 0.532 m/s²
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude of the acceleration of the box in terms of the friction force f?
Answer:
Explanation:
From Newton's second Law of Motion,
F = ma
Ff + F = ma
Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.
Magnitude of the acceleration:
a = Ff+F/m
This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive