Answer:
a) U = 506 J, b) U = 37.11 J, c) U = 0
Explanation:
The gravitational power energy is given by the expression
U = m g (y -y₀)
In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0
Let's use trigonometry to find the child's height
h = y = L - L cos θ
we substitute
U = m g L (1 - cos θ)
a) when the chain is horizontal θ = 90 and cos 90 = 0
U = mg L
weight and mass are related
W = mg
m = W / g
U = 230 2.20
U = 506 J
b) θ = 33.0º
cos 33 = 0.83867
U = 230 (1 - 0.83867)
U = 37.11 J
c) in this case θ = 0 cos 0 = 1
U = 0
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
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]
Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente
Answer:
The mount saved is $ 0.105.
Explanation:
Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately.
The electrical energy is given by
E = P x t
where, P is the electrical power and t is the time.
When he is using the shower for 20 minutes, the energy consumed is
E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh
When he is using the shower for 15 minutes, the energy consumed is
E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh
The difference in energy is
E'' = E - E' = 1.4 - 1.05 = 0.35 kWh
The money saved is
= 4 0.3 x 0.35 = $ 0.105
A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.
Answer:
1568J
Explanation:
Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.
Use conservation of Energy
ΔUg+ΔKE=0
ΔUg= mgΔh=2*9.8*(20-100)=-1568J
ΔKE-1568J=0
ΔKE=1568J
since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J
Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 2.76 m/s, they then have the same kinetic energy. Calculate the original speeds of the two cars.
Let m be the mass of the second car, so the first car's mass is 2m.
Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.
Let u and v be the speeds of the first car and the second car, respectively. At the start,
• the first car has kinetic energy
K/2 = 1/2 (2m) u ² = mu ² ==> K = 2mu ²
• the second car starts with kinetic energy
K = 1/2 mv ²
It follows that
2mu ² = 1/2 mv ²
==> 4u ² = v ²
When their speeds are both increased by 2.76 m/s,
• the first car now has kinetic energy
1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²
• the second car now has kinetic energy
1/2 m (v + 2.76 m/s)²
These two kinetic energies are equal, so
m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²
==> 2 (u + 2.76 m/s)² = (v + 2.76 m/s)²
Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.
IS ANYONE THERE..??!
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmm y
MCQ
................
Answer:
I think it would be (-7 C )..
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?
Answer:
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Explanation:
The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:
[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)
Where:
[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.
[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.
[tex]\Delta P[/tex] - Pressure change, in pascals.
If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:
[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]
[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]
[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = [tex]\sqrt{2qV/m}[/tex]
we calculate
v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]
v = [tex]\sqrt{632.3353 \ 10^8}[/tex]
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v[tex]\frac{m v}{2 q r}[/tex]
we calculate
B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]
B = 1.1413 10⁻² T
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080 arcsec,as measured by the Hipparcos satellite. What is the distance to Betelgeuse, and what is the uncertainty in that measurement?
We have that the distance to Betelgeuse, and the uncertainty in that measurement is
[tex]d=(221.7\pm39.33)pc[/tex]Uncertainty U = 0.00080
From the Question we are told that
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080
Generally
[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]
Where
Parallax [tex]P =0.00451[/tex]
Uncertainty [tex]U = 0.00080[/tex]
Generally the equation for the distance is mathematically given as
[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]
Therefore
[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]
[tex]d=(221.7\pm39.33)pc[/tex]
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The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N
uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm
Answer:
0.113 J
Explanation:
Applying,
w = ke²/2................. Equation 1
Where w = workdone in stretching the spring, k = spring constant, e = extension
make k the subject of the equation
k = 2w/e²................ Equation 2
From the question,
Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m
Substitute these values into equation 2
k = (2×3)/0.17²
k = 6/0.17
k = 35.29 N/m
(a) if the spring from 37 cm to 45 cm,
Then,
w = ke²/2
Given: e = 45-37 = 8 cm = 0.08
w = 35.29(0.08²)/2
w = 0.113 J
When Peter tosses an egg against a sagging sheet, the egg doesn't break due to
A) reduced impulse.
B) reduced momentum.
C) both of these
D) neither of these
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
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(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)
Answer:
t=0.417s
Explanation:
After the ball hits the racket it is in freefall(assume air resistance as negligible)
so a=-g
use
x-x0=v0t+1/2at^2
Plug in givens
20=50t-4.9t^2
Solve quadratic equation using quadratic formula
t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.535 T. If the coil carries a current of 26.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?
The object's kinetic energy changes according to
dK/dt = 15 J/s
If v is the object's initial speed, then its initial kinetic energy is
K (0) = 1/2 (5 kg) v ²
Use the fundamental theorem of calculus to solve for K as a function of time t :
[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]
After t = 13 s, the object's kinetic energy is
K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J
Put this as the left side in the equation above for K(t) and solve for v :
[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]
==> v ≈ 9.5 m/s
A cat's displacement is 15 meters to the right in 7.0 seconds. If, at the start of the 7.0 seconds, the cat was moving at a velocity of 2.0 m/s left what was its final velocity?
Answer:
6.3 m/s
Explanation:
From the given information:
The displacement (x) = 15 m
time (t) = 7.0 s
initial velocity = -2.0 m/s (since it is moving in the opposite direction)
We need to determine the acceleration then find the final velocity.
By applying the kinematics equation:
[tex]x = ut + \dfrac{1}{2}at^2[/tex]
[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]
Now, to determine the final velocity by using the equation:
v = u + at
v = -2 + 1.184(7.0)
v = 6.288 m/s
v ≅ 6.3 m/s
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball hits a wall that is 25.8 meters horizontally from where it started. How high (meters) does the ball hit on the wall?
6.07 m
Explanation:
Given:
[tex]v_0=24.5\:\text{m/s}[/tex]
[tex]\theta_0 = 35.5°[/tex]
First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that
[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]
or
[tex]t = 1.29\:\text{s}[/tex]
To find the vertical height where the ball hit the wall, we use
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]
[tex]\:\:\:\:=6.07\:\text{m}[/tex]
A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
How does the theory of relativity explain the gravity exerted by massive objects?
A. More massive objects create stronger forces of gravity.
B. More massive objects create shallower curves of space-time.
C. More massive objects pull objects from farther away.
D. More massive objects create larger curves of space-time.
(D)
Explanation:
The more massive an object is, the greater is the curvature that they produce on the space-time around it.
The theory of relativity explain the gravity exerted by massive objects is
more massive objects create larger curves of space-time (option-d).
Do bigger objects exert more gravity?The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.
Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.
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Upon completing an interview, it is important that you send a follow-up thank you
note/letter/e-mail because it will show that you are a person who appreciates an opportunity.
A True
B
False
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its angular acceleration.
Answer:
A = 26.875 rad/s²
Explanation:
Given the following data;
Initial angular speed, Uw = 150 rads/s.
Final angular speed, Vw = 580 rads/s.
Time = 16 seconds.
To calculate the angular acceleration;
From kinematics equation;
At = Vw - Uw
Where;
A is the angular acceleration.t is the timeVw is the final angular speed.Uw is the initial angular speed.Substituting into the formula, we have;
A*16 = 580 - 150
16A = 430
A = 430/16
A = 26.875 rad/s²
Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.
Diagram A
Diagram B
Diagram C
Diagram D
Answer:Diagram A
Explanation:
Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.
