Answer:
n × t_1/2
Exmplanation:
The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;
Age of meteorite= n × t_1/2
where;
n = quantity of the meteorite
t_/2 = half life of the meteorite
Thus:
The correct formula is; n × t_1/2
1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?
Answer:
1) attached below
2) assumption that the earth is spherical
Explanation:
1) Four illustrations of a globe
attached below
2) Reason for distortions at areas that do not fall on lines of tangency or secancy
The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
as the ball rises the vertical component of it's velocity_____. explain
Answer:
Decreases
Explanation:
because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
(a) How much mechanical energy was lost during the collision with the floor?
(b) A basketball player dribbles the ball from a height of 1.37 m by exerting a constant downward force on it for a distance of 0.132 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.37 m, what is the magnitude of the force?
Answer:
a)[tex]|\Delta E|=4.58\: J[/tex]
b)[tex]F=61.90\: N[/tex]
Explanation:
a)
We can use conservation of energy between these heights.
[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]
[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]
Therefore, the lost energy is:
[tex]|\Delta E|=4.58\: J[/tex]
b)
The force acting along the distance create a work, these work is equal to the potential energy.
[tex]W=\Delta E[/tex]
[tex]F*d=mgh[/tex]
Let's solve it for F.
[tex]F=\frac{mgh}{d}[/tex]
[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]
Therefore, the force is:
[tex]F=61.90\: N[/tex]
I hope is helps you!
A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is
Answer:
the weight of the air in pound-force (lb-f) is 325.8 lbf
Explanation:
Given;
dimension of the room, = 15 ft by 15 ft by 20 ft
density of air in the room, ρ = 0.0724 lbm/ft³
The volume of air in the room is calculated as;
Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³
The mass of the air is calculated as;
mass = density x volume
mass = 0.0724 lbm/ft³ x 4,500 ft³
mass = 325.8 lb-m
The weight of the air is calculated as;
Weight = mass x gravity
Weight = 325.8 lb-m x 32.174 ft/s²
Weight = 10482.29 lbm.ft/s²
The weight of the air in pound-force (lb-f) is calculated as;
1 lbf = 32.174 lbm.ft/s²
[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]
Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf
The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
a. F = m²wr
b. F = mwr²
c. F = mw²r
d. F = mwr.
Answer:
Correct option not indicated
Explanation:
There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).
The formula to calculate a centripetal force (F) is
F = mv²/r
Where m is mass, v is velocity and r is radius
where
While the formula to calculate a centrifugal force (F) is
F = mω²r
where m is mass, ω is angular velocity and r is radius of the circular path.
From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.
NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.
Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor
Answer:
560.54 Ω
Explanation:
Applying,
V = IR'............... Equation 1
Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors
make R' the subject of the equation
R' = V/I............ Equation 2
From the question,
Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A
Substitute these values into equation 2
R' = 5/(2.23×10⁻³ )
R' = 2242.15 Ω
Since the fours resistor are connected in series and they are equal,
Therefore the values of each resistor is
R = R'/4
R = 2242.15/4
R = 560.54 Ω
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
Define relative density.
Relative density is the ratio of the density of a substance to the density of a given material.
A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach
Answer: 11.36 m
Explanation:
Given
Mass of roller coaster is m=1200 kg
Initial speed of roller coaster is v=15 m/s
Energy at bottom and at the top is same i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]
Thus, the highest point reach by the roller coaster is 11.36 m
Answer:
11.36m
Explanation:
a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. assume that microwave energy is generated uniformly on the uipper surface. What is the power output of the oven
Complete question is;
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
Answer:
Power ≈ 600,000 W
Explanation:
We are given;
Frequency; f = 2400 Hz
height of the oven cavity; h = 25 cm = 0.25 m
base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²
total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J
We want to find the power output and we know that formula for power is;
P = workdone/time taken
Formula for time here is;
t = h/c
Where c is speed of light = 3 × 10^(8) m/s
Thus;
t = 0.25/(3 × 10^(8))
t = 8.333 × 10^(-10) s
Thus;
Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))
Power ≈ 600,000 W
If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference frame?
a. Yes, we would both view the motion from the same reference point because both of us are at rest in Earth’s frame of reference.
b. Yes, we would both view the motion from the same reference point because both of us are observing the motion from two points on the same straight line.
c. No, we would both view the motion from different reference points because motion is viewed from two different points; the reference frames are similar but not the same.
d. No, we would both view the motion from different reference points because response times may be different; so, the motion observed by both of us would be different.
Answer:
the correct is C
Explanation:
The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.
In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.
Consequently, the measurement carried out is related by
r = r’ + x
where the bold letters indicate wind blowers.
With these explanations we review the different answers, the correct one is C
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]
[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]
Therefore, the number of turns of wire needed is 573.8 turns
Three spheres (water, iron and ice) of the exact same volume are submerged in a tub of water. After the spheres are lined up, they are released. The spheres are made of plastic with the same density as water, ice, and iron.
Required:
a. Compare the weights of the three spheres.
b. Compare the buoyant forces on the three spheres.
c. What direction does the net force push on each of the spheres?
d. What happens to each sphere after it is released?
Answer:
(a) Iron > plastic > ice
(b) Same on all
(c) Iron downwards, plastic net force zero, ice upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Explanation:
Three spheres have same volume , plastic, ice and iron.
(a) The weight is given by
Weight = mass x gravity = volume x density x gravity
As the density of iron is maximum and the density of ice is least so the order of the weight is
Weight of iron > weight of plastic > weight of ice
(b) Buoyant force is given by
Buoyant force = Volume immersed x density of fluid x g
As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.
(c) Net force is
F = weight - buoyant force
So, the net force on the iron sphere is downwards
On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces
Answer:
The answer is C. Isolated System
Answer:
C. Isolated system
Explanation :
∵According to law of conservation of momentum ,In an isolated system ,the total momentum remains conserved.
convert 56km/h to m/s.
Explanation:
15.556 metres per second
Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances
Answer:
I = 0.0857 A
Explanation:
Given that,
Power consumed by the cellphone, P = 300 mW
The voltage of the battery, V = 3.5 V
Let I is the current flowing through the cell-phone. We know that,
P = VI
Where
I is the current
So,
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]
So, the current flowing the cell-phone is 0.0857 A.
find out the odd one and give reason (length, volume, time, mass
Answer:
Time
Explanation:
The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
Assume that I = E/(R + r), prove that 1/1 = R/E + r/E
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
[tex]I = \frac{ E}{ R + r} \\[/tex]
[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]
Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have
[tex]➺ \: I \: (R + r) = E \times 1[/tex]
[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]
[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]
[tex]\boxed{ Hence\:proved. }[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
How are elastic and inelastic collisions different?
A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.
B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.
C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.
D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.
Answer:
a
Explanation:
Answer:
the answer is c
'
Explanation:
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
The angular velocity of an object is given by the following equation: ω(t)=(5rads3)t2\omega\left(t\right)=\left(5\frac{rad}{s^3}\right)t^2ω(t)=(5s3rad)t2 What is the angular displacement of the object (in rad) between t = 2 s and t = 4 s?
Answer:
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
Explanation:
The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:
[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)
Where [tex]t[/tex] is the time, measured in seconds.
The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:
[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)
Then we proceed to integrate on the function in time:
[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]
[tex]\Delta \theta = 20\,rad[/tex]
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
