You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the same height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
A) Before A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration
B) Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A
C) The two blocks hit the ground at the same time, because they both ackelerate at the same rate while falling. The greater gravitational force on block B is compersated giving it the same acceleration
D) The two blocks hit the ground at the same time because they move at the same constant speed while falling.The force on the blocks are irrelevent once they start falling
ok so im not that good when it comes to physics but i think its C)
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.
What is the gravitational potential energy of a ball of mass 2.00 kg which is tossed to a height of 13.0 m above the ground? Answer in J, taking the potential energy to be 0.00 J at the ground.
Answer:
I believe the answer is 254.8 J, or rounded 255 J.
Explanation:
The formula for potential energy is:
PE=m(h)g
This means the mass (m) times height (h) times gravity (m). Gravity is 9.8 m/s (meters per second). Putting all of the numbers into it would equal:
PE=2(13)9.8
This equals 254.8 exactly, or if the assignment calls for you to round, 255.
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choices remains the same, but the apparent brightness is decreased by a factor of two. is decreased by a factor of four, and the apparent brightness is decreased by a factor of four. is decreased by a factor of four, but the apparent brightness remains the same. is decreased by a factor of two, and the apparent brightness is decreased by a factor of two. remains the same, but the apparent brightness is decreased by a factor of four.
Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
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
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 .
A boat has a mass of 7660 kg. Its engines generate a drive force of 4080 N due west, while the wind exerts a force of 680 N due east and the water exerts a resistive force of 1160 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign
Answer:
0.29 m/s due west.
Explanation:
According to newton's second law,
Net force acting on an object = mass×acceleration
From the question,
F+F₁+F₂ = ma................ Equation 1
Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.
Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg
substitute into equation 1
4080-680-1160 = 7660(a)
2240 = 7660a
Therefore,
a = 2440/7660
a = 0.29 m/s due west.
The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal law of gravitation states that the magnitude of the gravitational force between two masses M1 and M2 separated by a distance R is given by the following equation:________.
F = G (M1 x M2) / R2
G = 6.67 x 10-11 Nm2/kg2
a. Calculate the value of the gravitational force between an electron (mass = 9.11 x 10-31 kg) and a proton (mass is 1836 times greater than the mass of an electron) if the two particles are separated by 3.602 nanometers. (1 nanometer or 1 nm = 1 x 10-9 m)
F= ______ N
b. The force created in the above question is:
1. repulsive
2. attractive
Answer:
a.[tex]F=7.83\times 10^{-51} N[/tex]
b.Attractive
Explanation:
We are given that
[tex]F=\frac{GM_1M_2}{R^2}[/tex]
[tex]G=6.67\times 10^{-11} Nm^2/kg^2[/tex]
Mass of an electron,[tex]M_1=9.11\times 10^{-31} kg[/tex]
Mass of proton,[tex]M_2=1836\times 9.11\times 10^{-31} kg[/tex]
Distance between electron and proton,R=[tex]3.602nm=3.602\times 10^{-9} m[/tex]
[tex]1nm=10^{-9} m[/tex]
a.Substitute the values then we get
[tex]F=\frac{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1836\times 9.11\times 10^{-31}}{(3.602\times 10^{-9})^2}[/tex]
[tex]F=7.83\times 10^{-51} N[/tex]
b.We know that like charges repel to each other and unlike charges attract to each other.
Proton and electron are unlike charges therefore, the force between proton and electron is attractive.
You are woken up in the morning on the farm by a rooster. The rooster is located 30 meters away on top of the 5 meter barn house. If you throw your alarm clock at an angle of 53 degrees at the rooster, so that it hits the rooster on the way up, with what velocity do you need to throw your alarm clock
Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = 18.73 \ m/s[/tex]
Explanation:
From the question we are told that
The distance of the rooster from farm house is [tex]d = 30 \ m[/tex]
The height of the rooster from the ground is [tex]h = 5 \ m[/tex]
The angle at which the alarm clock is thrown is [tex]\theta = 53 ^o[/tex]
Let the velocity at which the alarm clock is thrown be v
So the horizontal component of v is mathematically represented as
[tex]v_h = v cos \theta[/tex]
The distance covered by the alarm clock toward the horizontal direction at velocity v is
[tex]d = v_h[/tex]
=> [tex]d = v_h = vcos \theta * t[/tex]
=> [tex]vcos \theta * t = 30[/tex]
substituting for [tex]\theta[/tex]
[tex]v (cos (53)) * t = 30[/tex]
=> [tex]v * t = 50[/tex]
Considering the motion of the alarm clock in the vertical direction
So the vertical component of v is mathematically represented as
[tex]v_v = vsin \theta[/tex]
The distance covered in the vertical direction is mathematically evaluated as follows
From the equation of motion we have
[tex]h = vsin\theta * t - \frac{1}{2} g t^2[/tex]
[tex]v* t sin\theta - 4.9t^2 = 5[/tex]
Recall [tex]v * t = 50[/tex]
So
[tex]50 \ sin \theta -4.9t^2 = 5[/tex]
substituting for [tex]\theta[/tex]
[tex]50 \ sin (53) -4.9t^2 = 5[/tex]
=> [tex]t =\sqrt{7.1289}[/tex]
[tex]t =2.673 \ s[/tex]
Now
[tex]v * t = 50[/tex]
So
[tex]v * 2.673 = 50[/tex]
=> [tex]v = 18.73 \ m/s[/tex]
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 64.7 N, Jill pulls with 61.5 N in the northeast direction, and Jane pulls to the southeast with 171 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey. What is the direction of the net force? Express this as the angle from the east direction between 0.
Answer:
F = 241.8 N θ = -18.7
Explanation:
To find the net force, we use Newton's second law, for this we decompose the force using trigonomer
Force 2
F2 = 61.5 N
cos 45 = F₂ₓ / F2 F₂ₓ = F2 cos 45
sin 45 = F₂_y / F2 F₂_y = F2 sin 45
force 3
F3 = 171 N
cos 45 = F₃ₓ / F3
sin 45 = F₃_y / F3
the total force can be found with its components
axis x direction (East-West)
Fₓ = 64.7 +61.5 cos 45 + 171 cos 45
Fₓ = 229.1 N
Y axis (direction North -Sur)
F_y = 61.5 sin 45 - 171 sin 45
F_y = - 77,428 N
the resulting force is
F = Fx i ^ + Fy j ^
F = (229.1 i⁻ 77,428 j⁾ N
we can use the Pythagorean theorem to find the module
F = √ (229.1 2 + 77,428 2)
F = 241.8 N
let's use trigonometry to find the direction
tan θ = F_y / Fₓ
θ = tan⁻¹ (F_y / Fₓ)
θ = tan⁻¹ (-77,428 / 229.1)
θ = -18.7
A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Answer:
The velocity is [tex]v = 8.85 m/s[/tex]
Explanation:
From the question we are told that
The mass of the skier is [tex]m_s = 60 \ kg[/tex]
The initial speed is [tex]u = 4.0 \ m/s[/tex]
The height is [tex]h = 10 \ m[/tex]
According to the law of energy conservation
[tex]PE_t + KE_t = KE_b + PE_b[/tex]
Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as
[tex]PE_t = mg h[/tex]
substituting values
[tex]PE_t = 60 * 4*9.8[/tex]
[tex]PE_t = 2352 \ J[/tex]
And [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill
And [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as
[tex]KE_b = 0.5 * m * v^2[/tex]
substituting values
[tex]KE_b = 0.5 * 60 * v^2[/tex]
=> [tex]KE_b = 30 v^2[/tex]
Where v is the velocity at the bottom
And [tex]PE_b[/tex] is the potential energy at the bottom which equal to zero due to the fact that height is zero at the bottom of the hill
So
[tex]30 v^2 = 2352[/tex]
=> [tex]v^2 = \frac{2352}{30}[/tex]
=> [tex]v = \sqrt{ \frac{2352}{30}}[/tex]
[tex]v = 8.85 m/s[/tex]
Answer:
The Skier's velocity at the bottom of the hill will be 18m/s
Explanation:
This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.
That is
[tex]mgh = 0.5 mv^{2}[/tex]
solving for velocity, we will have
[tex]v= \sqrt{2gh}[/tex]
hence her velocity will be
[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]
This is the velocity she gains from the slope.
Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s
The Skier's velocity at the bottom of the hill will be 18m/s
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.
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
(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!
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:
In Fig. on the right, what is the acceleration at 1.0 s?
Answer:
10 m/s²
Explanation:
Acceleration is the slope of the velocity vs time graph.
At t = 1.0 s, the slope of the line is:
a = Δv/Δt
a = (20 m/s) / (2.0 s)
a = 10 m/s²
Answer:
10m/s2
Explanation:
Acceleration is defined as change in velocity over the time period
Now at 1second the velocity is 10m/s.
Hence the acceleration is V-U/t;
Where V is velocity at 10s and U is velocity when the object is at rest.
So we have;
Acceleration = 10-0/ 1 = 10m/s2
Light from a fluorescent lamp is observed through a cloud of cool nitrogen gas. Again, two students are having a discussion about the kind of spectra that they would see.
Student 1: We would see absorption line spectra and the missing lines would correspond to the light from the fluorescent lamp.
Student 2: I disagree. We would see an emission line spectrum corresponding to nitrogen. This would happen because the nitrogen gas would absorb some energy from the fluorescent lamp and would reemit this energy which would result in an emission line spectrum.
Which student, if any, do you agree with and why?
Answer:
From the previous explanation Student No. 1 has the correct explanation
Explanation:
When the fluorescent lamp emits a light it has the shape of its emission spectrum, this light collides with the atoms of Nitrogen and excites it, so these wavelengths disappear, lacking in the spectrum seen by the observed, for which we would see an absorption spectrum
The nitrogen that was exited after a short time is given away in its emission lines, in general there are many lines, so the excitation energy is divided between the different emission lines, which must be weak
From the previous explanation Student No. 1 has the correct explanation
What real-world examples show not work being done? Can you think of examples other than resisting the force of gravity?
Answer:
Work is defined as a force doing a movement, for example, if with a force F we move an object a distance D, the work done is:
W = F*D
(note, the force is causing the movement, the product here is a dot product, this means that if the force and the displacement are perpendiculars, the product is zero)
So the examples where there are not work being done may be:
We do not have any movement:
For example, you can go to a wall in your house and push it really hard.
There is a force, but the wall will not move, so we have D = 0
and W = F*D = F*0 = 0
Because we have no motion.
Another case is where the force and the direction of motion are perpendiculars.
If we have for example a car, moving at a constant speed, and you push it sidewise (perpendicular to the direction of movement) we have a force being applied and movement, but those are in different directions (so the force does not cause the movement) so we dont have work being done.
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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How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?
Answer:
b
Explanation:
The pulley has an efficiency of 78%. How much useful work does the pulley do if you perform 775 J of work on the pulley
Answer:
605 J
Explanation:
Efficiency = work out / work in
0.78 = W / 775 J
W = 605 J
A stationary submarine using sonar emits a 1080 Hz sound wave that reflects off of an object moving towards the sub. The reflected sound is mixed with the 1080 Hz sound and a beat frequency of 80 Hz is observed. The speed of sound in water is 1400 m/s. How fast is the object moving
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
A crate sits on the ground. You push as hard as you can on it, but you cannot move it. At any given time when you are pushing, what is the magnitude of the static frictional force exerted on the crate compared to the magnitude of your push
Answer:
Equal magnitude
Explanation:
Static friction is the friction that exists between the stationary crate and the ground on which it's resting. static friction force is variable and depends on the external forces acting on an object. The higher the external force the higher the static friction until the crate starts moving when kinetic friction takes over.
For a stationary crate;
Static friction = external force exerted on the crate
Fₛ ≤ μₛ×N
Fₛ(max) ≤ μₛ×N
Where;
Fₛ = static friction force
μₛ = coefficient of static friction
N = weight of crate
Therefore,the magnitude of the static frictional force exerted on the stationary crate must be equal to the magnitude of your push.
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 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
A body moves due north with velocity 40 m/s. A force is applied
on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes
some time for that change and what is the direction of applied
external force?
Answer:
the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Explanation:
The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:
[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]
p: momentum
m: mass
[tex]\Delta v[/tex]: change in the velocity
The sign of the change in the velocity determines the direction of rate of change. Then you have:
[tex]\Delta v=v_2-v_1[/tex]
v2: final velocity = 35m/s
v1: initial velocity = 40m/s
[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]
Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
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].
Water is to be boiled at sea level (1 atm pressure) in a 30-cm-diameter stainless steel pan placed on top of a 18-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water. (Round the final answer to three decimal places.)
Answer:
Explanation:
18 kW = 18000 J /s
60% of 18kW = 10800 J/s
Latent heat of evaporation of water
= 2260 x 10³ J / kg
kg of water being evaporated per second
= 10800 / 2260 x 10³ kg /s
= 4.7787 x 10⁻³ kg / s
= 4.78 gm / s .
wich of the following are commonly distributed by veterinary assistants in typical veterinary practice?
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