Answer:
v = 103.70 m/s
Explanation:
To find the speed of the object, you first calculate the frequency of the reflected wave, by suing the information about the beat frequency:
[tex]f_b=|f_1-f_2|[/tex] (1)
fb: frequency of the beat = 80Hz
f1: frequency of the submarine generated by the submarine = 1080Hz
By solving the equation (1) you have that f2 can take two values:
[tex]f_2=1080Hz-80Hz=1000Hz\\f_2=1080Hz+80Hz=1160Hz[/tex]
You use the second value (1160Hz) because the reflected wave comes from an object that is moving toward the sub.
Next, you use the formula for the Doppler effect's, for an object that is getting closer:
[tex]f'=f(\frac{v_w}{v_w-v_s})[/tex] (2)
f': perceived frequency = 1160 Hz
f: frequency of the source = 1080 Hz
vw: speed of sound in water = 1400 m/s
vs: speed of the source = ?
You solve the equation (2) for vs, and you replace the values of the rest of the parameters:
[tex](v_w-v_s)f'=fv_w\\\\v_s=\frac{v_w(f'-f)}{f}\\\\v_s=\frac{(1400m/s)(1160Hz-1080Hz)}{1080Hz}\\\\v_s=103.70\frac{m}{s}[/tex]
hence, the speed of the object that is moving toward the sub is 103.70 m/s
Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Answer:
The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons
Explanation:
Binding energy can be given by the formula:
Binding energy = Binding energy per nucleon * Number of nucleons
This means that if the binding energy per nucleon = x MeV
Where x is a positive real number
The nucleus with 60 nucleons will have Binding energy = 60x MeV
The nucleus with 200 nucleons will have binding energy = 200x MeV
For a +ve x, 200x > 60x
Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.
Binding energy
Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.
It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.
Thus the response above is correct.
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А
mass exerts force of
5.6 X 10^-10N on
another mass
when
seperated 93cm apart. If
one mass is the square root of
the other
Find
value of
the two masses.
Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor
Answer:
μ = 0.336
Explanation:
We will work on this exercise with the expressions of transactional and rotational equilibrium.
Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation
fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0
fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0
fr = cotan θ (W / 2 + 0,3 W_painter)
Now let's write the equilibrium translation equation
X axis
F1 - fr = 0
F1 = fr
the friction force has the expression
fr = μ N
Y Axis
N - W - W_painter = 0
N = W + W_painter
we substitute
fr = μ (W + W_painter)
we substitute in the endowment equilibrium equation
μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)
μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)
we substitute the values they give
μ = cotan θ (12/2 + 0.3 55) / (12 + 55)
μ = cotan θ (22.5 / 67)
μ = cotan tea (0.336)
To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45
cotan 45 = 1 / tan 45 = 1
the result is
μ = 0.336
A 35.0-kg child swings in a swing supported by two chains, each 2.96 m long. The tension in each chain at the lowest point is 436 N. (a) Find the child's speed at the lowest point. Consider all the vertical components of force acting on the swing when it is at its lowest point and relate them to the acceleration of the swing at that instant. m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)
Answer:
6.69m/s,529N
Explanation:
Now we have 3 forces acting on the boy at its least point, the two tensions on the string and its weight. The tensions are acting upwards why its weight is acting downwards.
Hence the net force causes the child to swing in a circular fashion without skidding off the swing.
This net force is the centripetal force.
The weight of the child is mass × g
35×9.8=343N
The net force = 436+436-343 = 529N
The centripetal force is defined mathematically as;
F = mass × velocity square/ length of string.
Hence 529 = 35V^2/2.96
V^2 = 529×2.96/35 =44.7383
V=√44.7383 =6.69m/s
B. The force exerted by the seat on the child is the net force which keeps the boy from falling.
529N
Please help in the 2nd question
Answer:
[tex]q=4\times 10^{-16}\ C[/tex]
Explanation:
It is given that,
The charge on an object is 2500 e.
We need to find how many coulombs in the object. The charge remains quantized. It says that :
q = ne
[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]
So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000
Consider a situation in which you are moving two point charges such that the potential energy between them decreases. (NOTE: ignore gravity).
This means that you are moving the charges:
a) Closer to each other
b) Farther apart
c) Either A or B
Answer: Option A
Explanation:
The potential energy decreases in the case when the charges are opposite and they attract each other.
In this case there is no external energy required in order to put the charges together.
This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.
Example: a negative charge and a positive charge.
Two identical point charges q=71.0 pCq=71.0 pC are separated in vacuum by a distance of 2d=29.0 cm.2d=29.0 cm. Calculate the total electric flux ΦΦ through the infinite surface placed at a distance dd from each charge, perpendicular to the line on which the point charges are located.
Answer:
The electric flux at the infinite surface is ZERO
Explanation:
From the question we are told that
The point charge are identical and the value is [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]
The distance of separation is [tex]D = 29.0 \ cm = 0.29 \ m[/tex]
The distance of both from the infinite surface is d
Generally the electric force exerted by each of the charge on the infinite surface is
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO
You have two charges; Q1 and Q2, and you move Q1 such that the potential experienced by Q2 due to Q1 increases.
Gravity should be ignored.
Then, you must be:
a) Moving Q1 further away from Q2.
b) Moving in the opposite direction to that of the field due to Q1
c) Moving Q1 closer to Q2.
d) Moving in the same direction as the field due to Q1.
e) Any of the above
Given that,
First charge = Q₁
Second charge = Q₂
The potential experienced by Q2 due to Q1 increases
We know that,
The electrostatic force between two charges is defined as
[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]
Where,
k = electrostatic constant
[tex]Q_{1}[/tex]= first charge
[tex]Q_{2}[/tex]= second charge
r = distance
According to given data,
The potential experienced by Q₂ due to Q₁ increases.
We know that,
The potential is defined from coulomb's law
[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]
[tex]V\propto\dfrac{1}{r}[/tex]
If r decrease then V will be increases.
If V decrease then r will be increases.
Since, V is increases then r will decreases that is moving Q₁ closer to Q₂.
Hence, Moving Q₁ closer to Q₂.
(c) is correct option.
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point charge of 3.65×10−6 C3.65×10−6 C is placed at the center of the shell. What is the electric field magnitude EE a distance 0.795 m0.795 m from the center of the spherical shell?
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
[tex]\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}[/tex]
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
[tex]Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C[/tex]
Finally, you obtain for E:
[tex]E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}[/tex]
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
The angular velocity of a flywheel obeys the equation ωz(t)=A+Bt2, where t is in seconds and A and B are constants having numerical values 2.65 (for A) and 1.60 (for B).
Required:
a. What are the units of A and B if ωz is in rad/s?
b. What is the angular acceleration of the wheel at (i) t = 0 and (ii) t = 5.00 s?
c. Through what angle does the flywheel turn during the first 2.00 s?
Answer:
Explanation:
a )
Iz( t ) = A + B t²
Iz( t ) = angular velocity
putting dimensional formula
T⁻¹ = A + Bt²
A = T⁻¹
unit of A is rad s⁻¹
BT² = T⁻¹
B = T⁻³
unit of B is rad s⁻³
b )
Iz( t ) = A + B t²
dIz / dt = 2Bt
angular acceleration = 2Bt
at t = 0
angular acceleration = 0
at t = 5
angular acceleration = 2 x 1.6 x 5
= 16 rad / s²
Iz( t ) = A + B t²
dθ / dt = A + B t²
integrating ,
θ = At + B t³ / 3
when t = 0 , θ = 0
when t = 2
θ = At + B t³ / 3
= 2.65 x 2 + 1.6 x 2³ / 3
= 5.3 + 4.27
= 9.57 rad .
Flywheel turns by 9.57 rad during first 2 s .
(a) The unit of A is rad/s and the unit of B is rad/s³.
(b) The angular acceleration of the flywheel at 0 s is 0 and at 5 s is 16 rad/s²
(c) The angular displacement of the flywheel during the first 2 seconds, is 9.57 rad.
The given parameters;
z(t) = A + Bt²A = 2.65 and B = 1.6The units of A and B if z(t) is in radian per second (rad/s), are calculated as follows;
[tex]z(t)[\frac{rad}{s} ] = A[\frac{rad}{s} ] \ + \ Bt^2[\frac{rad}{s^3} ][/tex]
Thus, the unit of A is rad/s and the unit of B is rad/s³.
The angular acceleration of the flywheel is calculated as follows;
[tex]a = \frac{d\omega }{dt} =2Bt[/tex]
when, t = 0
a = 2(1.6)(0) = 0
when t = 5 s
a = 2(1.6)(5) = 16 rad/s²
The angular displacement of the flywheel during the first 2 seconds, is calculated as follows;
[tex]\theta = \int\limits {z(t)} \, dt\\\\\theta = At \ + \ \frac{Bt^3}{3} \\\\when, \ t = 2\ s;\\\\\theta = (2.65\times 2) \ + \ (\frac{1.6\times 2^3}{3} )\\\\\theta = 9.57 \ rad[/tex]
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Which statement best explains why the overall charge on an atom is zero?
ОООО
The positive charge of the neutrons in the nucleus equals the negative charge in the electron cloud.
The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
The negative charge of the neutrons in the nucleus equals the positive charge in the electron cloud.
The negative charge of the protons in the nucleus equals the positive charge in the electron cloud.
Answer:
B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
Explanation:
For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.
Answer:
B
Explanation:
This is described by Gauss a scientist.
The positive charge is found in the proton in the nucleus.
The neutron has no charge.
The positive charge radiates in all directions and a counter negative charge ensues.
In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10−3s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball?
Answer:
J = 8.4 kg*m/s
Explanation:
The magnitude of the impulse, J, on the ball can be calculated using the following equation:
[tex] J = F*t [/tex] (1)
Where:
F: is the force = 12000 N
t: is the time = 0.70x10⁻³ s
So, by entering the values above into equation (1) we can find the impulse on the ball:
[tex] J = F*t = 12000 N*0.70 \cdot 10^{-3} s = 8.4 kg*m/s [/tex]
Therefore, the impulse on the ball is 8.4 kg*m/s.
I hope it helps you!
The magnitude J of the impulse on the ball is 8.4 kg m/s.
Calculation of the impulse:Since it applies a force of 12,000 N to the ball for a time of 0.70×10−3s.
So,
The magnitude is
[tex]= Force \times time\\\\= 12,000 \times 0.70\times 10^{-3}[/tex]
= 8.4 kg m/s
Here we basically multiplied the force with the time to determine the magnitude.
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A cab driver heads south with a steady speed of V,22.0 m/s tor t 3.00 min, then makes a right turn and travels at v225.0 m/s for t2-2.40 min, and then drives northwest at v 30.0 m/s for 1.00 min. For this 6.40-min trip, calculate the following. Assume +x is In the eastward direction
(a) total vector displacement (Enter the magnitude in n and the direction in degrees south of west.) direvlion o south of west
(b) average speed (in m/s) m/s
(c) average velocity (Enter the magnitude in m/s and the direction in degrees south of west.) rmagnitude direction Im/'s o south of west
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return
Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
I was having trouble with this physics question
Worth 15 points
Answer:
68.79 N, 13.84° N of W
Explanation:
The law of cosines can be used to find the magnitude of the sum. F1 is 30° N of W, and F2 is 30° S of W, so the exterior angle of the force triangle is 30°+20° = 50°. The interior angle is the supplement of that. The angle between F1 and F2 in the force triangle representing the sum is 130°, so the sum of forces is ...
|F|^2 = |F1|^2 +|F2|^2 -2·|F1|·|F2|·cos(130°)
= 50^2 +25^2 -2·50·25·cos(130°) ≈ 4731.969
|F| ≈ √4731.969 ≈ 68.79 . . . . newtons
The angle α between F and F1 can be found from the law of sines.
sin(α)/|F2| = sin(130°)/|F|
α = arcsin(|F2|/|F|·sin(130°)) ≈ 16.16°
The diagram shows this to be the angle south of F1, so the angle of the sum vector F is 30° -16.16° N of W = 13.84° N of W.
The resultant force vector is 68.79 N at an angle of 13.84° N of W.
Answer:
68.98N, 13.8° N of W
Explanation:
The Forces F1+ F2 = 50N 30°N of W + 25N 20°S of W.
This forces can be split into horizontal and vertical components and are sum as such.
The horizontal and vertical component of F1 are;
50N cos 30= 43.30N W
50N sin 30 = 25N
The horizontal and vertical component of F2 are;
25N cos20°=23.49N West
25N sin20°=8.55N South
Sum of horizontal forces =43.30N+23.49 = 66.99N
Sum of vertical forces =25-8.55=16.45N{North is at the positive side of the y axis and South at the negative side}
The resultant sum of this forces is √(66.99)^2 + (16.45)^2=√4758.26=68.98N
The angle at which this force moves is Tan^{-1} 16.45/66.99 = 13.8° N of W
The Force therefore is 68.98N, 13.8° N of W
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
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Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller elements, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor
Answer:
Element C will be best for a nuclear fission reaction
Explanation:
Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle of 40° clockwise from the x axis. Find out resultant vector with its magnitude and direction using components method.
Answer:
The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
Explanation:
To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.
For M vector you obtain:
[tex]M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}[/tex]
For N vector:
[tex]N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}[/tex]
The resultant vector is the sum of the components of M and N:
[tex]F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}[/tex]
The magnitude of the resultant vector is:
[tex]|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm[/tex]
And the direction of the vector is:
[tex]\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°[/tex]
hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
brianna swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2s. What is the tangential speed of the ball
Answer:
4.3 m/s
Explanation:
What is the impulse on a car (750 kg) that accelerates from rest to 5.0 m/s in 10 seconds
Explanation:
impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)
Answer:
3750Ns
Explanation:
Impulse is defined as Force × time
Force = mass × acceleration,
Hence impulse is;
mass × acceleration × time.
From Newton's second law
Force × time = mass × ∆velocity
750× 5 = 3750Ns
∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.
0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distance of 2 m
Answer:
1.6 J
Explanation:
Work = change in energy
W = ΔKE
Fd = KE
(0.8 N) (2 m) = KE
KE = 1.6 J
f a curve with a radius of 97 m is properly banked for a car traveling 75 km/h , what must be the coefficient of static friction for a car not to skid when traveling at 100 km/h ?
Answer:
The coefficient of static friction is 0.26
Explanation:
Given;
radius of the road, R = 97 m
banking velocity, V₁ = 75 km/h = 20.83 m/s
velocity of the moving car, V₂ = 100 km/h = 27.78 m/s
Car in a banked circular turn:
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} )[/tex]
where;
θ is the angle between the horizontal ground and road in which the car move on
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} ) \\\\\theta = tan^{-1}(\frac{20.83^2}{9.8*97} ) \\\\\theta = 24.5^o[/tex]
During this type of motion, the body acquires some acceleration which tends to retain the circular motion towards its center, known as centripetal acceleration.
There are two components of this acceleration;
Parallel acceleration, [tex]a_|_|[/tex] = a*Cosθ
Perpendicular acceleration, a⊥ = a * Sinθ
Parallel acceleration, [tex]a_|_|[/tex] [tex]= \frac{V^2*Cos \theta}{R}[/tex]
Perpendicular acceleration, a⊥ [tex]= \frac{V^2*Sin \theta}{R}[/tex]
Apply Newton's second law of motion;
sum of perpendicular forces acting on the car;
ma⊥ [tex]= F_N - mg*cos \theta[/tex]
[tex]m(\frac{V^2*Sin \theta}{R} ) = F_N - mg*Cos \theta\\\\F_N = mg*Cos \theta + m(\frac{V^2*Sin \theta}{R} )[/tex] --------equation (1)
sum of parallel forces acting on the car
m[tex]a_|_|[/tex] [tex]= mg*Sin \theta - F_s[/tex]
[tex]m(\frac{V^2*Cos \theta}{R} ) = mg*Sin \theta - F_s\\\\F_s = mg*Sin \theta - m(\frac{V^2*Cos \theta}{R} )[/tex] ---------equation (2)
Coefficient of static friction is given as;
[tex]\mu = \frac{F_s}{F_N}[/tex]
Thus, divide equation (2) by equation (1)
[tex]\frac{F_s}{F_N} = \frac{mg*Sin \theta - m(\frac{V^2*Cos \theta}{R}) }{mg*Cos \theta + m(\frac{V^2*Sin \theta}{R}) } \\\\\frac{F_s}{F_N} = \frac{g*Sin \theta - (\frac{V^2*Cos \theta}{R}) }{g*Cos \theta + (\frac{V^2*Sin \theta}{R}) }[/tex]
V = V₂ = 27.78 m/s
θ = 24.5°
R = 97 m
g = 9.8 m/s²
Substitute in these values and solve for μ
[tex]\frac{F_s}{F_N} = \frac{9.8*Sin(24.5)+ (\frac{27.78^2*Cos (24.5)}{97}) }{9.8*Cos (24.5) + (\frac{27.78^2*Sin (24.5)}{97}) }\\\\\frac{F_s}{F_N} = \frac{4.0641 \ - \ 7.2391}{8.91702 \ + \ 3.299} = -0.26\\\\| \mu| = 0.26[/tex]
Therefore, the coefficient of static friction is 0.26
Suppose you are given the following equation, where xf and xi represent positions at two instants of time, vxi is a velocity, ax is an acceleration, t is an instant of time, and a, b, and c are integers. xf = xita + vxitb + ½axtc.
Required:
For what values of a, b, and c is this equation dimensionally correct?
Answer:
Explanation:
xf = xita + vxitb + ½axtc.
xf is displacement , dimensional formula L .
Xi initial displacement , dimensional formula L
t is time , dimensional formula T ,
vxi is velocity , dimensional formula LT⁻¹
ax is acceleration , dimensional formula = LT⁻²
xf = xi t a + vxi t b + ½ ax t c.
L = aLT + b LT⁻¹ T + c LT⁻² T
From the law of uniformity , dimensional formula of each term of RHS must be equal to term on LHS
aLT = L
a = T⁻¹
b LT⁻¹ T = L
b = 1 ( constant )
c LT⁻² T = L
c = T
so a = T⁻¹ , b = constant and c = T .
Which of the following is NOT a type of electromagnetic wave?
Seismic waves
Visible light
Radio waves
Microwave
Answer:
Seismic waves
Explanation:
seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.
Answer:
Siesmic waves
Explanation:
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s.
Answer:
a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s
c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s
Explanation:
This exercise is about kinematics in two dimensions
a) find the position of the particle on each axis
X axis
Since there is no acceleration on this axis, we can use the relation of uniform motion
v = x / t
x = v t
we substitute
x = 6.00 t
Y Axis
on this axis there is an acceleration and there is no initial speed
y = v₀ t + ½ a t²
y = ½ at t²
we substitute
y = ½ 1.00 t²
y = 0.500 t²
in vector position is
d = x i ^ + y j ^
d = (6.00 t i ^ + 0.500 t²) m
b) x axis
as there is no relate speed is concatenating
vₓ = v₀
vₓ = 6.00 m / s
y Axis
there is an acceleration and the initial speed is zero
[tex]v_{y}[/tex] = v₀ + a t
v_{y} = a t
v_{y} = 1.00 t
the velocity vector is
v = vₓ i ^ + v_{y} j ^
v = (6.00 i ^ + 1.00 t j ^) m / s
c) the coordinates for t = 4 s
d = (6.00 4 i ^ + 0.50 4 2 j⁾
d = (24.00 i ^ + 8.00 j^ ) m
x = 24.0 m
y = 8.00 m
d) the velocity of for t = 4 s
v = (6 i ^ + 1 5 j ^)
v = (6.00 i ^ + 5 j^ ) m/s