Answer:
Linear mass density,[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]
Explanation:
Given that,
Mass of the string, m = 0.3 g = 0.0003 kg
Length of the string, l = 1.5 m
The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :
[tex]\lambda=\dfrac{m}{l}[/tex]
[tex]\lambda=\dfrac{0.0003\ kg}{1.5\ m}[/tex]
[tex]\lambda=0.0002\ kg/m[/tex]
or
[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]
So, the linear mass density of a string is [tex]2\times 10^{-4}\ kg/m[/tex]. Hence, this is the required solution.
To calculate the linear mass density of the string, you divide the mass in kilograms by the length in meters. For a 0.3g and 1.5m string, this yields a linear mass density of 0.0002 kg/m.
The student asks how to calculate the linear mass density (μ) of a string. The linear mass density is defined as mass per unit length. To calculate it for a given string, you divide the mass of the string by its length. The provided string has a mass of 0.3g, which should be converted to kilograms (0.0003 kg), and a length of 1.5m.
The formula to calculate linear mass density is:
μ = mass/length.
Therefore, the linear mass density of the string is:
μ = 0.0003 kg / 1.5 m = 0.0002 kg/m.
A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) Is the force attractive or repulsive?
Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive
Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:
[tex]F=\frac{k*q1*q2}{d^{2}}[/tex]
Replacing the dat we obtain F=82 nN.
The force is repulsive because the points charged have the same sign.
Problem 6: A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 94 m/s2 for t1 = 23 s. The first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 39 m/s2 for the time interval t2 = 34 s. Part (a) Enter an expression for the rocket's speed v1 at time t1 in terms of the variables provided. Part (b) Enter an expression for the rocket s speed v2 at the end of the second period of acceleration in terms of the variables provided in the problem statement Part (c) Using vow expressions for speeds v1 and v2 calculate the total distance trailed in meters, by the rocket from launch until the end of the second period of acceleration.
Answer:
a) v1 = a1*t1 = 2162 m/s
b) v2 = v1 + a2*(t2-t1) = 2591 m/s
c) [tex]Dt = D1 + D2 = \frac{v1^{2}}{2*a1} + \frac{v2^{2}-v1^{2}}{2*a2}=51004.5m[/tex]
Explanation:
For any movement with constant acceleration:
Vf = vo + a*Δt. Replacing the propper values, with get the answers for parts a) and b):
v1 = a1*t1 = 2162 m/s
v2 = v1 + a2*(t2-t1) = 2591 m/s
Using the formula for displacement we can calculate the total distance asked on part c):
[tex]V_{f}^{2}=V_{o}^2+2*a*D[/tex] Solving for D and replacing the values for each part of the launch:
[tex]D=\frac{V_{f}^{2}-V_{o}^{2}}{2*a}[/tex]
D1 = 24863m
D2 = 26141.5m
Finally we add D1 + D2 for the total distance:
D = 51004.5m
The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2. reduce the distance to R/4. double the charge to 2Q. reduce the distance to R/2. double the distance to 2R.
Answer:
Double the charge to 2Q
Explanation:
The magnitude of a electric field caused by a charge Q at a distance R can be find by the following expression:
[tex]E = K\frac{Q}{R^2}[/tex]
Where K is the Coulomb constant.
As the relationship between the charge and the electric fiel is proportional, by simply doubling the charge, you would double the magnitude of the electric field.
Notice that the distance affects the magnitude of the electric field by the inverse square. So if you half the distance, the magnitude of the field will quadruple.
To double the electric field's magnitude at point P from a point charge Q at distance R, double the charge to 2Q while keeping the distance R unchanged. Option B is the correct option.
To double the magnitude of the electric field at point P due to a point charge Q a distance R away from P, you can follow Coulomb's law. Electric field (E) is directly proportional to the charge (Q) and inversely proportional to the square of the distance (R) between the point charge and the observation point.
If you double the charge to 2Q while keeping the distance R the same, the new electric field magnitude at point P will be double the original E. This is because E ∝ Q. Other options like reducing the distance to R/2 or R/4, or doubling the distance to 2R, would not result in a doubling of the field magnitude but rather lead to different field strengths according to the inverse square law, and altering both the charge and distance simultaneously is not necessary to achieve this specific goal. Option B is the correct option.
To learn more about charge ,visit this link:
https://brainly.com/question/34758769
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The appropriate question is :
The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2 OR
A) reduce the distance to R/4.
B) double the charge to 2Q.
C)reduce the distance to R/2.
D) double the distance to 2R.
f the electric field is zero at a particular point, must the electric potential be zero at the same point? Explain
Answer:
No
Explanation:
As we know that the electric field nullity does not define that the electric potential will be zero at that point.
For example consider the two positive charge at the mid point of these charge electric field is zero but potential is finite.The electric potential has two contribution means it is positive if charges are positive and it is negative if charges are negative.A ball is dropped from rest at the top of a 6.10 m
tallbuilding, falls straight downward and collides inelastically
withthe ground, and bounces back. The ball loses 10% of itskinetic
energy every time it collides with the ground. Howmany bounces can
the ball make and still reach a windowsill that is2.38 m above the
ground?
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]
and
[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]
[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]
So on
[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]
[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
The ball can bounce approximately 8 times before it reaches the height of 2.38 meters after accounting for a 10% energy loss each collision.
This problem involves a ball undergoing inelastic collisions, losing 10% of its kinetic energy with each bounce, and determining how many times it can bounce to still reach a windowsill 2.38 m high.
First, let's calculate the initial potential energy (PEinitial) of the ball when it is dropped from a height (hinitial) of 6.10 m:
PEinitial = mghinitial
where, g = 9.8 m/s² (acceleration due to gravity)
As it falls, this entire potential energy converts into kinetic energy (KEinitial) at the ground:
KEinitial = PEinitial = mghinitial
Upon each bounce, the ball loses 10% of its kinetic energy. Therefore, it retains 90% of its kinetic energy:
KEnew = 0.9 × KEprevious
To find out how high it can bounce after each loss of energy, convert kinetic energy back into potential energy:
PE = KE = mgh
After each bounce, the height the ball can reach is calculated by applying the 90% retention factor:
hnew = 0.9 × hprevious
Starting with h0 = 6.10 m:
Next height (h1) = 0.9 × 6.10 m = 5.49 mNext height (h2) = 0.9 × 5.49 m = 4.94 mNext height (h3) = 0.9 × 4.94 m = 4.45 mContinue this process until hn is less than 2.38 m:The number of bounces can be calculated using the formula: hn = 6.10 × (0.9)n, where n is the number of bounces. Set hn = 2.38 m and solve for n:
2.38 = 6.10 × (0.9)n
Divide both sides by 6.10:
0.39 ≈ (0.9)n
Taking the natural log (ln) of both sides:
ln(0.39) = n × ln(0.9)
Finally, solving for n:
n ≈ ln(0.39) / ln(0.9) ≈ 8
Therefore, the ball can bounce approximately 8 times and still reach the windowsill that is 2.38 m above the ground.
A student is walking with a constant speed of ????????1meters per second along High Street and sees a puddle ????????1 meters ahead of her. A bus driver is driving parallel to the student along High Street as well. At the moment the bus is ????????2meters behind the student, the bus driver decides he wants to splash he student with water by driving over the puddle as the student walks past. Determine the expression for the speed that the bus must have in terms of the given variables such that the bus and the student reach the puddle at the same time to splash the student.
Answer:
The expression is[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex]
The speed must be 3 m/sExplanation:
We know that speed is:
[tex]v= \frac{distance}{time}[/tex].
So, to find the speed for the bus, we need to know:
a. How far the bus is from the puddle.b. In how much time will the student reach the puddle.Lets call [tex]v_{student}[/tex] the speed of the student, and [tex]d_{student \ to \ puddle}[/tex] the distance from the student to the puddle.
We can obtain the time taking
[tex]v_{student}= \frac {d_student \ to \ puddle}{t}[/tex]
as t must be the time that the student will take to reach the puddle:
[tex]t= \frac {d_{student \ to \ puddle}}{v_{student}}[/tex]
The bus is at a distance [tex]d_{bus \ to \ the \ student}[/tex] behind the student, so, the total distance that the bus must travel to the puddle is:
[tex]d_{bus \ to \ the \ puddle} = d_{bus \ to \ the \ student} + d_{student \ to \ puddle}[/tex]
Taking all this togethes, the formula must be:
[tex]v= \frac{d_{bus \ to \ the \ puddle}}{t}[/tex].
[tex]v= \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{\frac {d_{student \ to \ puddle}}{v_{student}}}[/tex].
[tex]v= v_{student} \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{d_{student \ to \ puddle}}[/tex].
[tex]v= v_{student} (\frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} + \frac{d_{student \ to \ puddle}}{d_{student \ to \ puddle}} )[/tex].
[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex].
And this is the formula we are looking for.
Taking the values from the problem, we find
[tex]v= 1 \frac{m}{s} ( 1 + \frac{2 m}{1 m})[/tex]
[tex]v= 1 \frac{m}{s} ( 1 + 2)[/tex]
[tex]v= 3 \frac{m}{s}[/tex]
The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°. The resulting acceleration has a magnitude of 20 m/s^2. What is the mass of the body? O 2.4kg O 2.2kg O 2.7kg O 3.1kg
Answer:
b) 2.2 kg
Explanation:
Net force acting on an object is the sum of the two forces acting on the body.
The net force is calculated using the parallelogram law of vectors.
F =[tex]\sqrt{{A^{2}} + B^{2}+2 A B cos \theta}[/tex]
Here A = 20 N , B = 35 N and θ =80°
Net Force = F = 43.22 N
Acceleration = a = 20 m/s/s
Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket’s height when it runs out of fuel; (b) find the rocket’s velocity when it runs out of fuel; (c) find the maximum height the rocket reaches; (d) find the rocket’s velocity the instant before the rocket crashes into the ground; and (e) find the total elapsed time from launch to ground impact.
Answer:
a) 200 m
b) 100 m/s
c) 709 m
d) -118.2 m/s
e) 26.24 s
Explanation:
The rocket flies upward with constant acceleretion.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
V0 = 0
Y(t) = 1/2 * 25 * t^2
Y(t) = 12.5 * t^2
And speed under constant acceleration:
Vy(t) = Vy0 + a * t
Vy(t) = 25 * t
It burns for 4 s and runs out of fuel
Y(4) = 12.5 * 4^2 = 200 m
V(4) = 25 * 4 = 100 m/s
Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity
It will be under constant acceleration. These new equations will have different starting constants.
Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2
Vy(t) = Vy4 + g * (t - 4)
When it reaches its highest point it will have a speed of zero.
0 = Vy4 + g * (t - 4)
0 = 100 - 9.81 * (t - 4)
100 = 9.81 * (t - 4)
t - 4 = 100 / 9.81
t = 10.2 + 4 = 14.2 s
At that moment it will have a height of:
Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m
The rocket will fall and hit the ground:
Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2
0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)
0 = -4.9 * t^2 + 139.2 * t -278.4
Solving this equation electronically:
t = 26.24 s
At that time the speed will be:
Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s
An electron with speed v0 = 5.08×10^6 m/s is traveling parallel to an electric field of magnitude E = 9100 N/C . Part A How far will the electron travel before it stops? Express your answer to three significant figures and include the appropriate units.
Part B How much time will elapse before it returns to its starting point? Express your answer to three significant figures and include the appropriate units.
Answer:
Part A: [tex]8.06\times10^{-3}\ m[/tex]
Part B: [tex]6.36\times 10^{-9}\ s[/tex]
Explanation:
Given:
[tex]v_o[/tex] = initial velocity of the electron = [tex]5.08\times 10^{6}\ m/s[/tex][tex]E[/tex] = electric field strength = 9100 N/CAssumptions:
[tex]m[/tex] = mass of the electron = [tex]9.1\times 10^{-31}\ kg[/tex][tex]v[/tex] = final velocity of the electron = 0 m/s[tex]x[/tex] = distance at which the electron comes to rest[tex]e[/tex] = magnitude of charge on an electron = [tex]1.6\times 10^{-19}\ C[/tex][tex]t[/tex] = taken by the electron to return to its initial position[tex]s[/tex] = displacement of the electronPart A:
Since the electron moves in the direction of the electric field, the electric force will act on it in the direction opposite to electric field. This electric force does work on it to make the electron come to rest.
Using the work-energy theorem, the work done by the electric field will be equal to the kinetic energy change of the electron.
[tex]\therefore -eEx = \dfrac{1}{2}m(v^2-v_o^2)\\\Rightarrow -eEx=-\dfrac{1}{2}mv_o^2\\\Rightarrow x=\dfrac{mv_o^2}{2eE}\\\Rightarrow x=\dfrac{9.1\times 10^{-31}\times (5.08\times 10^{6})^2}{2\times 1.6\times10^{-19}\times 9100}\\\Rightarrow x=8.06\times 10^{-3}\ m[/tex]
Hence, the electron comes to rest by travelling a distance of [tex]8.06\times 10^{-3}\ m[/tex].
Part B:
In this part, let us first find out the acceleration of the electron due to the electric force.
[tex]a = -\dfrac{eE}{m}\\\Rightarrow a= -\dfrac{1.6\times10^{-19}\times 9100}{9.1\times 10^{-31}}\\\Rightarrow a= -1.6\times 10^{15}\ m/s^2\\[/tex]
The electron moves with the above acceleration constantly as it moves in the uniform electric field.
Since the electron is supposed to move from a point and then again move back to the same point. This means the displacement of the electron is zero.
[tex]i.e.,\ s=0\\\Rightarrow v_ot+\dfrac{1}{2}at^2=0\\\Rightarrow (v_o+\dfrac{1}{2}at)t=0\\\Rightarrow \dfrac{(2v_o+at)}{2}t=0\\\Rightarrow t = 0\,\,\, or\,\,\, (2v_o+at)=0\\\Rightarrow t = 0\,\,\, or\,\,\, t=\dfrac{-2v_o}{a}\\[/tex]
Since the electron starts moving at t = 0 s.
[tex]\therefore t = \dfrac{-2v_o}{a}\\\Rightarrow t=\dfrac{-2\times 5.08\times 10^6}{-1.6\times 10^{15}}\\\Rightarrow t= 6.36\times 10^{-9}\ s[/tex]
Hence, the electron returns to the starting position after [tex]6.36\times 10^{-9}\ s[/tex].
The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against gravity, what is the change in height of the center of mass?
Answer:
h = 4271.43 m
Explanation:
given,
Volume of the water = 1 m³
temperature decrease by = 10°C
heat removed from water
Q = m c ΔT
Q = ρ V c ΔT
= 1000 × 1 × 4186 × 10
= 4.186 × 10⁷ J
energy is used to do work to move the water against its weight
Q = force × displacement
4.186 × 10⁷ J = m g × h
4.186 × 10⁷ J = 1000 × 1 × 9.8 × h
h = 4271.43 m
hence, the change in height of is equal to h = 4271.43 m
A person is standing on a level floor. His head,
uppertorso, arms, and hands together weigh 438 N and have a center
ofgravity that is 1.28 m above the floor. His upper legs weigh 144
Nand have a center of gravity that is 0.760 m above the
floor.Finally, his lower legs and feet together weigh 87 N and
havea center of gravity that is 0.250 m above the floor. Relative
tothe floor, find the location of the center of gravity for
theentire body.
Answer:
The location of the center of gravity for the entire body, relative to the floor is 1.03 m
Explanation:
To find the center of gravity of a system of particles, we use that
[tex]R_{cog} = \frac{r_{1}*m_{1}+r_{2}*m_{2}+...+r_{n}*m_{n}}{M}[/tex]
where R is the vector center of gravity of the system, formed by n particles, and n masses.
In this case, for a person standing on the floor and being their body divided in three sectors, each one with a weight and an altitud in a specific point (center of gravity of the body sector), instead of mass we have every "particle" weight in Newtons (force instead of mass), being each "particle" in the formula, a sector of the body.
On the other hand, we use only magnitude for the calculation, because the gravity force is vertical to the floor, so instead of our vector formula, we use it in the vertical direction as a magnitude formula. Thus
[tex]Y_{cog} = \frac{y_{1}*W_{1}+y_{2}*W_{2}+y_{3}*W_{3}}{Wt}[/tex]
where Y is the center of gravity, y=1, 2, 3 is every "sector point" altitude from the floor, W=1, 2, 3 is every weight of a body "sector", and Wt is the sum of the three weights.
In this way we replace in our formula with the correspondent values
[tex]Y_{cog} = \frac{1.28m*438N+0.76m*144N+0.25m*87N}{669N}[/tex]
obtaining our result
[tex]Y_{cog}=1.03 m[/tex]
At t =0 one toy car is set rolling on a straight track with intial position 17.0 cm , intial velocity -3 cm/s, and constant acceleration 2.30 cm/s^2 . At the same moment , another toy car is set rolling on an adjacent track with initial position 9.5 cm , intial velocity 5.0 cm/s, and constant zero acceleration. (A) at the time, if any, do the two cars have equal speeds? (B) what are their speeds at that time? (c) at what time(s) , if any , do the cars pass each other? (D) what are their location at that time?
Answer:
a) 5.65 s
b) 5cm/s
c) They will pass each other at both 1.1168 s and 5.84s
d)15.084cm and 38.7 cm
Explanation:
For part A, you need to keep in mind that acceleration is the rate of change of velocity per unit of time. For a constant acceleration, this can be told in this way:
[tex]a = \frac{v - v_o}{t}[/tex]
Reordering this equation, we can get v in terms of the initial velocity, the acceleration, and the time elapsed:
[tex]v = at + v_o[/tex]
Now, we can get the expressions for velocity of each toy car, and equalize them:
[tex]v_1 =a_1t + v_o_1\\v_2 =a_2t + v_o_2\\v_1 = v2\\a_1t +v_o_1 =a_2t + v_o_2\\(a_1 - a_2)t = v_o_2 - v_o_1\\t = \frac{v_o_2 - v_o_1}{a_1 - a_2} = \frac{5cm/s - (-3cm/s)}{2.3 cm/s^2 - 0 cm/s^2}= 3.47 s[/tex]
As toy car has no acceleration and, therefore, constant speed, both car will have the same speed when toy car 1 reaches this velocity = 5cm/s
c) The position of car 1, as it follows a constant acceleration motion, is given by this equation:
[tex]x_1 = \frac{1}{2}a_1t^2 + v_o_1t + x_o_1[/tex]
The position for car 2, as it has constant velocity, is given by this equation:
[tex]x_2 = v_2t + x_o_2[/tex]
We equalize both equation to find the time where the cars pass each other:
[tex]x_1 = x_2\\\frac{1}{2}a_1t^2 + v_o_1t + x_o_1 = v_2t+x_o_2\\\frac{1}{2} a_1t^2 + (v_o_1 - v_2)t + x_o_1 - x_o_2 = 0\\\frac{1}{2}2.3m/s^2t^2 +(-3cm/s-5cm/s)t+ 17cm - 9.5cm = 0\\1.15t^2 -8t + 7.5 = 0 | a = 1.15, b = -8, c = 7.5\\t = \frac{-b +-\sqrt{b^2 - 4ac}}{2a} = 5.84s | 1.1168 s[/tex]
The car will pass each other at both 1.1168s and 5.84s.
For the positions, we solve any of the position equation with the solutions:
[tex]x = v_2*t + x_o_2 = 5cm/s *5.84s + 9.5cm = 38.7 cm\\x = 5cm/s * 1.1168s + 9.5cm = 15.084 cm[/tex]
The two toy cars have equal speeds at approximately t = 3.48s, with both traveling at 5 cm/s. They pass each other at t ≈ 1.99s, with their location at approximately 15.4 cm.
Part A: Equal Speeds:
We know the first car has an initial velocity of -3 cm/s and an acceleration of 2.30 cm/s². Its velocity at any time t can be given by v1 = v0 + at, that is -3 + 2.30t.
The second car has a constant velocity of 5.0 cm/s.
For them to have equal speeds, v1 = v2,
so -3 + 2.30t = 5.
Solving for t gives t ≈ 3.48 s.
Part B: Speeds at Equal Times:
The common speed when both cars have equal speeds can be found by substituting the time back into either equation for velocity.
Doing so for the first car gives
-3 + 2.30(3.48) ≈ 5 cm/s.
Part C: Cars Passing Each Other:
The cars pass each other when their positions are equal. Using the equations for position x1 = x01 + v01t + (1/2)at² and x2 = x02 + v02t,
we get 17 + (-3)t + (1/2)(2.30)t² = 9.5 + (5)t.
Solving for t gives two possible times, but only the positive one is relevant, t ≈ 1.99 s.
Part D: Location at Passing Time:
Substituting the time when they pass each other back into the position equations, for the first car we have
17 + (-3)(1.99) + (1/2)(2.30)(1.99)² ≈ 15.4 cm.
The second car will be at the same position since they are passing each other.
An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular acceleration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.
Answer:
0.06 Nm
Explanation:
mass of object, m = 3 kg
radius of gyration, k = 0.2 m
angular acceleration, α = 0.5 rad/s^2
Moment of inertia of the object
[tex]I = mK^{2}[/tex]
I = 3 x 0.2 x 0.2 = 0.12 kg m^2
The relaton between the torque and teh moment off inertia is
τ = I α
Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia
τ = 0.12 x 0.5 = 0.06 Nm
A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 lbf/in.2, ν1 = 1.25 ft3/lb and ends with p2 = 60 lbf/in.2, ν2 = 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n.
Answer:
V1=2.5ft3
V2=1ft3
n=1.51
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=2lb*1.25ft3/lb=2.5ft3
state 2
V2=m.v2
V2=2lb*0.5ft3/lb= 1ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/60)/ ln (1/2.5)
n=1.51
The car starts from rest and accelerates with an acceleration of 2 m/s^2 for 5 s. It then travels at a constant speed for 20 s, before decelerating at -5 m/s^2 until it reaches zero speed. How much distance did the car cover during this journey.
Answer:
Total distance covered during the journey is 235 m
Solution:
As per the question:
Initial velocity, v = 0 m/s
Acceleration, a = [tex]2 m/s^{2}[/tex]
Time, t = 5 s
Now,
For this, we use eqn 2 of motion:
[tex]d = vt + \farc{1}{2}at^{2}[/tex]
[tex]d = 0.t + \farc{1}{2}\times 2\times 5^{2} = 25 m[/tex]
The final speed of car after t = 5 s is given by:
v' = v + at
v' = 0 + 2(5) = 10 m/s
Now, the car travels at constant speed of 10 m/s for t' = 20 s with a = 0:
[tex]d' = vt + \farc{1}{2}at^{2}[/tex]
[tex]d' = 10\times 20 + \farc{1}{2}\times 0\times 20^{2} = 25 m[/tex]
d' = 200 m
Now, the car accelerates at a= - 5 [tex]m/s^{2}[/tex] until its final speed, v" = 0 m/s:
[tex]v"^{2} = v'^{2} + 2ad"[/tex]
[tex]0 = {10}^{2} + 2\times (- 5)d"[/tex]
[tex]100 = 10d"[/tex]
d" = 10 m
Total distance covered = d + d' + d" = 25 + 200 + 10 = 235 m
A non conducting sphere of radius 0.04 m has a charge of 5.0 × 10^-9 C deposited on it. CalculateThe magnitude of the electric field at 0.02m from the center of the sphere
Answer:
The electric field at a distance r = 0.02 m is 14062.5 N/C.
Solution:
Refer to fig 1.
As per the question:
Radius of sphere, R = 0.04 m
Charge, Q = [tex]5.0\times 10^{- 9} C[/tex]
Distance from the center at which electric field is to be calculated, r = 0.02 m
Now,
According to Gauss' law:
[tex]E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}[/tex]
Now, the charge enclosed at a distance r is given by volume charge density:
[tex]\rho = \frac{Q_{enclosed}}{area}[/tex]
[tex]\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}[/tex]
Also, the charge enclosed Q' at a distance r is given by volume charge density:
[tex]\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Since, the sphere is no-conducting, Volume charge density will be constant:
Thus
[tex]\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Thus charge enclosed at r:
[tex]Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}[/tex]
Now, By using Gauss' Law, Electric field at r is given by:
[tex]4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}[/tex]
Thus
[tex]E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}[/tex]
[tex]E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}[/tex]
E = 14062.5 N/C
The gauge pressure in your car tires is 2.40 x 10^5 N/m^2 at a temperature of 35.0°C when you drive it onto a ferry boat to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°C? (Assume that their volume has not changed.)
Answer:
The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
Explanation:
Given that,
Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]
Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]
Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]
We need to calculate the gauge pressure P₂
Using relation pressure and temperature
[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]
[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]
[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]
Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
Find the volume of a cylinder of height 10 cm if its base is a square of side 4 cm
Answer:
[tex]Volume=160 cm^3[/tex]
Explanation:
The volume of the cylinder is given by the area of the base multiplied by the height.
In this case:
Height:
[tex]h = 10 cm\\[/tex]
Base area = area of the square ([tex]area=side*side=side^2[/tex])
the side of the square is:
[tex]side=4cm[/tex]
thus, the area of the base:
[tex]area=(4cm)^2 = 16 cm^2[/tex]
Now we multiply this quantities, to find the volume:
[tex]Volume= 16 cm^2*10cm=160 cm^3[/tex]
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1 microsecond is 1x10 65). (Express your answers in vector form.) Part 1 (a) What is the average velocity of the electron? Vavg = < > m/s Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Part 2 (b) If the electron continues to travel at this average velocity, where will it be in another 9 us? 7 = < > m
Final answer:
The electron's average velocity is found to be (0 m/s, 316,000 m/s, -146,000 m/s), and after another 9 microseconds, it will be at the position (0.02 m, 4.464 m, -2.104 m).
Explanation:
To calculate the average velocity of an electron, we use the formula:
Vavg = (rf - ri) / Δt, where rf is the final position, ri is the initial position, and Δt is the time interval between the positions.
Given the initial position (0.02 m, 0.04 m, -0.06 m) and the final position (0.02 m, 1.62 m, -0.79 m), with a time difference of 5 microseconds (μs), which is 5 x 10-6 seconds:
The position change in vector form is
Δr = (0.02 m - 0.02 m, 1.62 m - 0.04 m, -0.79 m - (-0.06 m))
= (0 m, 1.58 m, -0.73 m).
Thus, the average velocity is
Vavg = Δr / Δt
= (0 m, 1.58 m, -0.73 m) / (5 x 10-6 s)
= (0 m/s, 316,000 m/s, -146,000 m/s)
The electron's new position after another 9 μs, moving with the same average velocity, is calculated by:
rnew = rf + Vavg × Δtnew
Here, Δtnew is 9 μs, which is 9 x 10-6 seconds, so:
rnew = (0.02 m, 1.62 m, -0.79 m) + (0 m/s, 316,000 m/s, -146,000 m/s) × (9 x 10-6 s)
= (0.02 m, 1.62 m + (2.844 m), -0.79 m - (1.314 m))
= (0.02 m, 4.464 m, -2.104 m).
When an object is thrown upwards and reaches its maximum height its speed is: a. Greater than the initial
b. Less than the initial
c. Same as the initial
d. Zero
Answer:
Option d
Explanation:
When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.
While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.
So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.
A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strikes the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed.
Answer:
Explanation:
Ball is thrown downward:
initial velocity, u = - 20 m/s (downward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(a) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(b) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = - 20 - 9.8 t
t = 2 second
Now the ball is thrown upwards:
initial velocity, u = 20 m/s (upward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(c) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(d) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = + 20 - 9.8 t
t = 6.09 second
Light travels at a speed of close to 3 x 10^5 km/s in vacuum. Given that it takes light 8 min and 19 s to travel the distance from the center of the Earth to the center of the Sun, how far away is the Sun from the Earth? (Astronomers use this as a "distance unit" called 1 Astronomical Unit or 1 au)
Answer:
1497×10⁵ km
Explanation:
Speed of light in vacuum = 3×10⁵ km/s
Time taken by the light of the Sun to reach the Earth = 8 min and 19 s
Converting to seconds we get
8×60+19 = 499 seconds
Distance = Speed × Time
[tex]\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km[/tex]
1 AU = 1497×10⁵ km
The Sun is 1497×10⁵ km from Earth
You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in the downward direction (assuming air resistance can be ignored). What was the average acceleration vector of the ball? Express your answer using two significant figures.
Answer:
The acceleration is -9.8 m/s²
Explanation:
Hi there!!
When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.
The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:
acceleration = final speed - initial speed/ elapsed time
acceleration = -4.3 m/s - 4.3 m/s / 0.88 s
acceleration = -9.8 m/s²
The average acceleration vector would be 0 m/s² due to the upward (negative) and downward (positive) accelerations cancelling. However, the average magnitude of acceleration regardless of direction is 9.8 m/s², which is simply the acceleration due to gravity.
Explanation:To answer your question about the average acceleration vector of a ball that was thrown upward, we can use the concept of free fall in physics. When you throw a ball upward, it initially slows down due to earth's gravitational pull until it stops at its highest point, then it starts accelerating downward due to gravity, until it reaches your hand again.
The acceleration when the ball is going upward will be the opposite of the acceleration when it's coming downward because they are in opposite directions but magnitude will be the same. We can assume the acceleration due to gravity as -9.8 m/s² (negative indicating upward direction) and when it's coming down it will be 9.8 m/s² (positive indicating downward direction).
So over the course of its flight (0.88s), the average acceleration would be ((-9.8)+(9.8))/2 = 0 m/s². However, this may not be what you're looking for as this is the averaged vector sum of the accelerations upward and downward.
If you are asking about the average magnitude of acceleration regardless of direction (in magnitude) it would be the average of the absolute values of the accelerations, which would be the gravitational acceleration g = 9.8 m/s².
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The fastest server in women's tennis is Sabine Lisicki, who recorded a serve of 131 mi/h (211 km/h) in 2014. Suppose that the acceleration of the ball was constant during the contact with the racket. Part A If her racket pushed on the ball for a distance of 0.15 m, what was the acceleration of the ball during her serve?
Using the formula for acceleration, a = (v^2) / 2d, and the given velocity and acceleration distance, the acceleration of the tennis ball during Lisicki's serve was approximately 11368 m/s². The force exerted by the racket is generally higher than the force due to gravity during this action.
Explanation:To calculate the acceleration of the tennis ball during Sabine Lisicki's serve, we can use the formula for acceleration: a = (v^2) / 2d, where 'v' corresponds to the final velocity and 'd' represents the distance over which the ball accelerated.
In this case, the final velocity 'v' is 211 km/h (converted to m/s gives us approximately 58.6 m/s), and Lisicki's racquet was in contact with the ball, causing it to accelerate over a distance 'd' of 0.15 m. Plugging these values into the formula gives us: a = (58.6 m/s)^2 / 2(0.15 m), which equals about 11368 m/s².
The average force exerted by the racket can be understood through the equation F = ma, a case of Newton's second law of motion. However, in this scenario, the force due to gravity is negligible as it's much smaller than the force exerted by the racket. The main focus here is the force exerted by the racket which made such a high acceleration possible.
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Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in this process?
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J
Answer:
(a) 992.87 g
(b) [tex]6.419\times 10^{14} J[/tex]
Solution:
As per the question:
Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g
(a) To calculate mass of He produced:
We know that:
Atomic mass of hydrogen is 1.00784 u
Also,
4 Hydrogen atoms constitutes 1 Helium atom
Mass of Helium formed after conversion:
[tex]4\times 1.00784 = 4.03136 u[/tex]
Also, we know that:
Atomic mass of Helium is 4.002602 u
The loss of mass during conversion is:
4.03136 - 4.002602 = 0.028758 u
Now,
Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]
Now,
For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g
Mass of He produced in the process:
[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]
(b) To calculate the amount of energy released:
We use Eintein' relation of mass-enegy equivalence:
[tex]E = M'c^{2}[/tex]
[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]
Earth orbits around the Sun at an average speed of 30 km/s. How far does the Earth move in a year (3.16 x 107 s) as it revolves around the Sun? Express your answer in kilometers using the correct number of significant figures
The Earth moves 9.48 × 10⁸ kilometers in a year as it revolves around the Sun, calculated by multiplying the average speed of 30 km/s by the time of one year, 3.16 × 10⁷ seconds.
Explanation:To calculate the distance that Earth moves in a year as it revolves around the Sun, we can use the formula for distance traveled, which is distance = speed × time. Given that Earth orbits around the Sun at an average speed of 30 kilometers per second (km/s) and the time it takes for one orbit is 3.16 × 107 seconds, the calculation would be:
30 km/s × 3.16 × 107 s = distance
Distance = 30 × 3.16 × 107 km
Distance = 94.8 × 107 km
Distance = 9.48 × 108 km
The Earth moves 9.48 × 108 kilometers in a year as it revolves around the Sun, using the correct number of significant figures.
A 70kh man jumping from a window lands in an elevated
firerescue net 11.0 m below the window. He momentarily stops when
hehas stretched the net by 1.50 m. Assuming that mechanicalenergy
is conserved during this process and the net functions likean ideal
spring, find the leastic potential energy of the net whenit is
stretched by 1.50 m.
Answer:
potential energy of net is 8.58 kJ
Explanation:
given data
mass m = 70 kg
height h = 11 m
stop distance x = 1.50 m
to find out
potential energy of net
solution
we know here man eventually stop
so elastic potential energy of net = change in potential energy of man body
so equation will be
potential energy of net = m×g×h ...................1
here m is mass and h is total height = ( h+ x)
height = 11 + 1.5 = 12.5 m and g is 9.8
so put here value in equation 1
potential energy of net = m×g×h
potential energy of net = 70×9.8×12.5
potential energy of net = 8575 J
so potential energy of net is 8.58 kJ
A massless spring is attached to a support at one end and has a 2.0 μC charge glued to the other end. A -4.0 μC charge is slowly brought near. The spring has stretched 1.2 cm when the charges are 2.2 cm apart. What is the spring constant of the spring? Express your answer with the appropriate units.
The spring constant of the spring is -1.376 N/m.
Explanation:To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the displacement of the spring is 1.2 cm, and the charges are 2.2 cm apart. The force exerted by the spring can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.
Given that the displacement (x) is 1.2 cm and the force exerted is caused by the electric force between the charges, which is given by Coulomb's Law, we can write the equation:
F = kx = k(1.2 cm) = (k/100 cm) * 1.2 cm = k/83.33
Similarly, the electric force between the charges is given by Coulomb's Law: F = kq1q2/r^2, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are 2.2 cm apart and have magnitudes of 2.0 μC and -4.0 μC, respectively.
When the charges are 2.2 cm apart, we can calculate the electric force:
F = k(2.0 μC)(-4.0 μC)/(2.2 cm)^2 = (-8.0 kμC^2)/(4.84 cm^2) = (-8.0 k)/(4.84 cm^2) μC^2
Equating the force exerted by the spring to the electric force, we have:
k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2
Solving for k:
k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2
k * (4.84 cm^2)/(83.33) = -8.0 k * μC^2
(4.84 cm^2)/(83.33) = -8.0 μC^2
k = (-8.0 μC^2) * (83.33)/(4.84 cm^2) = -137.6 μC^2/cm^2 = -1.376 N/m
Therefore, the spring constant of the spring is -1.376 N/m.
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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write R in vector form. (b) Write R showing the magnitude and direction in degrees.
Answer:
R = ( 4.831 m , 1.469 m ) Magnitude of R = 5.049 mDirection of R relative to the x axis= 16°54'33'Explanation:
Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude and θ.
So, for our vectors, we will have:
[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]
[tex]\vec{D}= ( 2.121 m , -2.121 m )[/tex]
and
[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]
[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]
Now, we can take the sum of the vectors
[tex]\vec{R} = \vec{D} + \vec{E}[/tex]
[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]
[tex]\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]
[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]
This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem
[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]
[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]
[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]
[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]
[tex]|\vec{R}| = 5.049 m[/tex]
To find the direction, we can use
[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]
[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]
[tex]\theta = arctan(0.304)[/tex]
[tex]\theta = 16\°54'33''[/tex]
As we are in the first quadrant, this is relative to the x axis.
Water has a mass per mole of 18.0 g/mol, and each water molecule (H20) has 10 electrons. (a) How many electrons are there in one liter (1.00 x 10 m ) of water?
Answer:
total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×[tex]10^{23}[/tex] molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 ×[tex]10^{23}[/tex] × 1
total number of electron in water is = 6.02×[tex]10^{24}[/tex] electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 × [tex]10^{-3}[/tex] m³
mass of 1 litter = 1000 × 1 × [tex]10^{-3}[/tex]
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × [tex]\frac{molecules per gram mole}{mass per mole}[/tex]
total number of electron in 1 litter = 1000 × [tex]\frac{6.02*10{24}}{18}[/tex]
total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron