Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
If the potential (relative to infinity) due to a point charge is V at a distance R from this charge, the distance at which the potential (relative to infinity) is 2V is
A. 4R
B. 2R
C. R/2.
D. R/4
Answer:
R/2
Explanation:
The potential at a distance r is given by :
[tex]V=\dfrac{kq}{r}[/tex]
Where
k is electrostatic constant
q is the charge
The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,
[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]
Put all the values,
[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]
So, the distance at which the potential (relative to infinity) is 2V is R/2.g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.
Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vertical height does the block reach above its starting point? Use the coefficients μk=0.20 andμs=0.50.
Answer:
The wood block reaches a height of 4.249 meters above its starting point.
Explanation:
The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]
[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]
[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)
Where:
[tex]h[/tex] - Maximum height of the wood block, in meters.
[tex]v[/tex] - Initial speed of the block, in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]m[/tex] - Mass, in kilograms.
[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.
[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.
If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:
[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]h = 4.249\,m[/tex]
The wood block reaches a height of 4.249 meters above its starting point.
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
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 309 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars
Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective
Answer:
Reflective
Explanation:
The radiation pressure of the wave that totally absorbed is given by;
[tex]P_{abs}= \frac{I}{C}[/tex]
and While the radiation pressure of the wave totally reflected is given by;
[tex]P_{ref}= \frac{2I}{C}[/tex]
Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.
g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?
Answer:
the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
Explanation:
Given the data in the question;
centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²
radius r₁ = 0.0189 m
radius r₂ = 0.0897 m
centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²
since the rotational period will be the same for the two disk,
we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.
a[tex]_c[/tex]₁ = (4π²r₁/T²)
make T² subject of formula
T² = 4π²r₁ / a[tex]_c[/tex]₁
we substitute
T² = ( 4 × π² × 0.0189 ) / 241
T² = 0.00309602528 s²
Now we use the same formula to find a[tex]_c[/tex]₂
a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )
we substitute
a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 ) / 0.00309602528
a[tex]_c[/tex]₂ = 1143.8 m/s²
Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
Question 8 of 10
What was the name of the book that Ibn al-Haytham wrote?
A. Weather and Air Flow
B. Book of Optics
C. Light and Vision
D. Book of Sound
Answer:b
Explanation:
A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
If the moon started it's orbit around the Earth from a spot in line with a certain star, it will return to that same spot in about _______.
Answer:
1 month
Explanation:
why kg is a fundamental unit?
This above answer helps a lot.
During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.
Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
Energy from the sun comes to Earth as radiant energy. Which of these is an example of radiant energy being converted to heat energy?
A Turning windmills transform mechanical energy into electrical energy.
B Black shirts feel hotter than light-colored shirts on a sunny day.
C Solar cells convert sunlight into electrical energy.
D Green plants use sunlight in photosynthesis.
Answer:
B
Explanation:
The radiant energy form the sun is absorbed by the black shirt and is converted to heat energy.
Answer:
B Black shirts feel hotter than light-colored shirts on a sunny day.
Explanation:
The energy from the sun also called solar energy is an energy source which reaches the earth as a form of radiant energy, that is it is transmitted without the movement of mass. Solar cells absorbs radiant energy from the sun into electrical energy for powering electrical devices.
During photosynthesis, sunlight absorbed by the chlorophyll of green plants is converted into chemical energy.
In black body, radiant energy abosrde are stored and converted to heat energy, reason dark colored clothes feels hotter than light colored on sunny days.
A 10.0kg of desk initial is pushed along a frictionless surface by a constant horizontal of force magnitude 12N Find the speed of the desk after it has moved through a horizontal distance of 5.0m
kylydljty many true dvx*&;'*+$_5+
A block of mass M is connected by a string and pulley to a hanging mass m.The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. Find the acceleration of the system and tensions on the string.
The free body diagram for the block of mass M consists of four forces:
• the block's weight, Mg, pointing downward
• the normal force of the table pushing upward on the block, also with magnitude Mg
• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left
• tension of magnitude T pulling the block to the right
For the block of mass m, there are only two forces:
• its weight, mg, pulling downward
• tension T pulling upward
The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.
Newton's second law gives us
T - 0.2Mg = Ma
mg - T = ma
where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :
mg - 0.2 Mg = (m + M) a
a = (m - 0.2M) / (m + M) g
a = 1.96 m/s²
Then the tension in the string is
T = m (g - a)
T = 78.4 N
g four small masses 0.2 kg each are connected by light rods 0.4m long to form a square.what is the moment of interia axis
Complete Question
Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.
Answer:
[tex]I=0.032kgm^2[/tex]
Explanation:
From the question we are told that:
Mass[tex]m=0.2kg[/tex]
Length [tex]l=0.4m[/tex]
Generally the equation for Inertia is mathematically given by
[tex]I=md^2[/tex]
[tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]
[tex]I=0.032kgm^2[/tex]
Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
is 6.0mm in diameter and lift a 1700kg object. Then, the wire elongates
0.17m. Calculate the Young’s modulus for the rope material.
Answer:
Young's modulus for the rope material is 20.8 MPa.
Explanation:
The Young's modulus is given by:
[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]
Where:
F: is the force applied on the wire
L₀: is the initial length of the wire = 3.1 m
A: is the cross-section area of the wire
ΔL: is the change in the length = 0.17 m
The cross-section area of the wire is given by the area of a circle:
[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]
Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:
[tex] F = T_{w} = W_{o} [/tex]
Where:
[tex] T_{w} [/tex]: is the tension of the wire
[tex]W_{o} [/tex]: is the weigh of the object = mg
m: is the mass of the object = 1700 kg
g: is the acceleration due to gravity = 9.81 m/s²
[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]
Hence, the Young's modulus is:
[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]
Therefore, Young's modulus for the rope material is 20.8 MPa.
I hope it helps you!
What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2
Answer:
810.94 m/s
Explanation:
Applying,
v = √(2gR)............. Equation 1
Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth
From the question,
Given: g = 0.636 m/s², R = 517 km = 517000 m
Substitute these values into equation 1
v = √(2×0.636×517000)
v = √(657624)
v = 810.94 m/s
Hence, the escape velocity is 810.94 m/s
how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?
Assuming no heat lost to the surrounding,
-5⁰C ice → 0⁰C ice
Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(2.0 x 10³) x (0-(-5))
Q = 50000J
0⁰C ice → 0⁰C water
Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg
Q = mLf
Q = 5(3.34 x 10⁵)
Q = 1670000J
0⁰C water → 100⁰C water
Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(4.2 x 10³) x (100-0)
Q = 2100000J
100⁰C water → 100⁰C steam
Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Q = mLv
Q = 5(2.26 x 10⁶)
Q = 11300000J
Total amount of heat required
= 50000 + 1670000 + 2100000 + 11300000
= 15120000J
Question 8 a-e plz
Answer:
(a) t = 0 s
(b) t = 0 s, 30 s, 55 s
(c) t = 40 s to t = 60 s
(d) t = 10 s to t = 15 s
(e) a = 6 m/s^2
Explanation:
(a) The car is at starting position at t = 0 s and v = 0 m/s.
(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.
(c) from t = 40 s to 60 s the car is moving in the negative direction.
(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.
(e) The slope of the velocity time graph gives acceleration.
a = (60 - 0) / (10 - 0) = 6 m/s^2
You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.
Answer:
the pressure exerted by the object is 392,000 N/m²
Explanation:
Given;
mass of the object, m = 8 kg
area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²
acceleration due to gravity, g = 9.8 m/s²
The pressure exerted by the object is calculated as;
[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]
Therefore, the pressure exerted by the object is 392,000 N/m²
A wave pulse travels along a stretched string at a speed of 200 cm/s. What will be the speed if:
a. The string's tension is doubled?
b. The string's mass is quadrupled (but its length is unchanged)?
c. The string's length is quadrupled (but its mass is unchanged)?
d. The string's mass and length are both quadrupled?
Answer:
a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s
Explanation:
The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.
So, v = √(T/μ)
v = √(T/m/l)
v = √(Tl/m)
a. The string's tension is doubled?
If the tension is doubled, T' = 2T the new speed is
v' = √(T'l/m)
v' = √(2Tl/m)
v' = √2(√Tl/m)
v' = √2v
v' = √2 × 200 cm/s
v' = 282.84 cm/s
b. The string's mass is quadrupled (but its length is unchanged)?
If the mass is quadrupled, m' = 4m the new speed is
v' = √(Tl/m')
v' = √(Tl/4m)
v' = (1/√4)(√Tl/m)
v' = v/2
v' = 200/2 cm/s
v' = 100 cm/s
c. The string's length is quadrupled (but its mass is unchanged)?
If the length is quadrupled, l' = 4l the new speed is
v' = √(Tl'/m)
v' = √(T(4l)/m)
v' = √4)(√Tl/m)
v' = 2v
v' = 200 × 2 cm/s
v' = 400 cm/s
d. The string's mass and length are both quadrupled?
If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is
v' = √(Tl'/m')
v' = √(T(4l)/4m)
v' = √(Tl/m)
v' = v
v' = 200 cm/s
Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.
Answer:
a) v₀ = 44.27 m / s, b) stone A v = 44.276 m / s, stone B v = 0.006 m / s
Explanation:
a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m
y = y₀ + v₀ t - ½ gt²
as the stone is released its initial velocity is zero
y- y₀ = 0 - ½ g t²
t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]
t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]
t = 4.518 s
now we can find the initial velocity of stone B to reach this height at the same time
y = y₀ + v₀ t - ½ g t²
stone B leaves the floor so its initial height is zero
100 = 0 + v₀ 4.518 - ½ 9.8 4.518²
100 = 4.518 v₀ - 100.02
v₀ = [tex]\frac{100-100.02}{4.518}[/tex]
v₀ = 44.27 m / s
b) the speed of the two stones at the meeting point
stone A
v = v₀ - gt
v = 0 - 9.8 4.518
v = 44.276 m / s
stone B
v = v₀ -g t
v = 44.27 - 9.8 4.518
v = 0.006 m / s
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)
A 9.0 V battery is connected across two resistors in series. If the resistors have resistances of and what is the voltage drop across the resistor?
Select one:
A. 4.6 V B. 9.4 V C. 8.6 V D. 4.4 V
Answer:
the answer to the question is known as D
