Given
A rectangular painting is 120 cm wide and 80 cm high.
To find
At what speed must the painting move parallel to its width if it is to appear to be square?
Explanation
According to length contraction, the proper length of an object contracts when it moves with a velocity.
Thus,
[tex]\begin{gathered} l=l_o\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow80=120\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow\frac{2}{3}=\sqrt{1-\frac{v^2}{c^2}} \\ \Rightarrow\frac{4}{9}=1-\frac{v^2}{c^2} \\ \Rightarrow\frac{v^2}{c^2}=1-\frac{4}{9} \\ \Rightarrow\frac{v^2}{c^2}=\frac{5}{9} \\ \Rightarrow v=\frac{\sqrt{5}}{3}c \end{gathered}[/tex]Conclusion:
The required velocity is
[tex]\frac{\sqrt{5}c}{3}\frac{m}{s}[/tex]In the graph of a bus his journey through town. Which of the following choices best represent where the bus is decelerating
Answer:
The bus is decelerating from B to C
Explanation:
The graph given shows the velocity of a bus at certain points in time. Higher the point on the graph, the higher the veocity. Now, the bus decelerates when its speed reduces from a higher value to a lower value. This we see as happening from point B to point C.
Hence, the bus decelerates from B to C.
At other points, you see the bus as either accelerating ( from O to A and D to E) or standing still (from A to B, C to D, and E to F). Only from B to C does the bus decelerate.
A ball player catches a ball 3.2 s after throwing it vertically upward.With what speed did he throw it? What height did it reach?Express your answers using two significant figures.
Since the time of flight of the ball is 3.2 seconds, that means for 1.6 seconds the ball was going upwards, then it reached the maximum height (where the velocity is zero), then went 1.6 seconds downwards, until it reaches the hand of the player again.
Using the formula below, we can find the initial velocity of the ball:
[tex]V=V_0+a\cdot t[/tex]Where V is the final velocity after t seconds, V0 is the initial velocity and 'a' is the acceleration.
Using V = 0, t = 1.6 s and a = -9.8 (gravity's acceleration), we have:
[tex]\begin{gathered} 0=V_0-9.8\cdot1.6\\ \\ V_0-15.68=0\\ \\ V_0=15.68\text{ m/s} \end{gathered}[/tex]Rounding the answer to two significant figures, we have an initial velocity of 16 m/s.
To find the maximum height, we can use the formula below:
[tex]\begin{gathered} \Delta S=V_0t+\frac{at^2}{2}\\ \\ \Delta S=15.68\cdot1.6-\frac{9.8\cdot1.6^2}{2}\\ \\ \Delta S=25.088-12.544\\ \\ \Delta S=12.544\text{ m} \end{gathered}[/tex]Rounding the answer to two significant figures, the maximum height is 13 m.
explain how the intensity of the UV light vaires across the Earth
Some factors determine the amount of UV radiation that reach certain part of Earth's surface. They are listed and briefly explained below:
• Cloud coverage. Water molecules on clouds scatter the radiation, hence the more clouds the less UV radiation.
,• Ozone. Similarly as cloud coverage, the more concentration of ozone the less UV radiation that reaches the surface of the Earth.
,• Angle of incidence. If the angle of incidence of the UV light is oblique the light will spread in a wider area, and hence the intensity is spread across this area.
,• Aerosols. The molecules of aerosols also scatter the UV light.
,• Elevation. The more the elevation the greater the amount of UV light.
1 pts
An coconut with a mass of 2 kg and a feather with a mass of 0.01 kg fall from a tree through the air to the ground
below, both eventually reaching terminal velocity. At terminal velocity, the amount of air-rresistance force what is the answer ? A grather on the coconut B the same on each C grather on the feather
Take into account that air-resistance force is greater against bodies with lower densities.
In this case, the feather has a lower density than the coconut, then, you can conclude that air-resistance force is greater on the feather (option C).
What is the relationship. Stern average kinetic energy of a gas and it’s temperature?
Average kinetic energy is directly proportional to its temperature-
E= f/2 N k T
E= energy
T= temperature
A bowling ball of mass 7.29 kg and radius 11.0 cm rolls without slipping down a lane at 3.00 m/s. Calculate the total kinetic energy.
We have the next information
m=7.29 kg
v=3 m/s
r=11cm=0.11m
We can find the kinetic energy using the next formula
[tex]KE=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]I is the moment of inertia
w is the angular velocity
First, we need to calculate the moment of inertia
[tex]I=\frac{2}{5}mr^2[/tex][tex]I=\frac{2}{5}(7.29)(0.11)^2=0.0353kgm^2[/tex]Then for the angular velocity
[tex]\omega=\frac{v}{r}[/tex][tex]\omega=\frac{3}{0.11}=27.28[/tex]then we will substitute the values in the kinetic energy formula
[tex]KE=\frac{1}{2}(7.29)(3)^2+\frac{1}{2}(0.0353)(27.28)^2[/tex][tex]KE=45.9J[/tex]The total kinetic energy is 45.9 J
i hve attached the question
The speed of the planet Venus 3.56 × 10⁵ m/s.
We are given that,
The radius of planet Venus = r = 1.08 × 10¹¹ m
The orbital time period = t = 225 days = 1.9 × 10⁶sec
The speed of planet Venus = v = ?
The formula of the time period of complete one revolution of planet to the sun is given as,
T = 2π/ω
ω = r.v
v = 2πr/T
Putting , the values in above equation we can get,
v = (2 × 3.14 × 1.08 × 10¹¹ m)/ ( 1.9 × 10⁶sec)
v = 3.56 × 10⁵ m/s
Therefore , the speed of planet Venus would be 3.56 × 10⁵ m/s
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A wheel Was spinning at 2.8 rad/s. It took 3.2 seconds to stop completely. What is the acceleration of the wheel?
Given
The angular velocity is
[tex]\omega=2.8\text{ rad/s}[/tex]The time taken,
[tex]t=3.2s[/tex]To find
The acceleration of the wheel
Explanation
The acceleration is
[tex]\begin{gathered} \alpha=\frac{\omega}{t} \\ \Rightarrow\alpha=\frac{2.8}{3.2}=\frac{0.875rad}{s^2} \end{gathered}[/tex]Conclusion
The acceleration is
[tex]0.875\text{ rad/s}^2[/tex]If the man and woman are taken to a planet where the acceleration due to gravity is twice that of earth repeat the woman mass was 25kg on earth and the man was 300N on another planet
The mass of the woman is 25 kg because the mass is constant.
The mass of the man can be found using the formula: W = mg, where g is double Earth's gravity.
[tex]\begin{gathered} m=\frac{W}{g} \\ m=\frac{300N}{2\cdot9.8\cdot\frac{m}{s^2}} \\ m=\frac{300}{19.6}kg \\ m\approx15.3\operatorname{kg} \end{gathered}[/tex]The mass of the man is 15.3 kg.
The weight of the man on Earth can be found with the same formula but using Earth's gravity.
[tex]\begin{gathered} W=15.3\operatorname{kg}\cdot9.8\cdot\frac{m}{s^2} \\ W=149.94N \end{gathered}[/tex]The weight of the man on Earth is 149.94 N.
At last, the weight of the woman on Earth can be found using the same method before.
[tex]\begin{gathered} W=25\operatorname{kg}\cdot9.8\cdot\frac{m}{s^2} \\ W=245N \end{gathered}[/tex]The weight of the woman on Earth is 245N.
I need help with this table pleasecalculate relative density of steel. Use table 3
Take into account that the relative density is given by:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}[/tex]where ρ, in this case, is the density of the steel and ρwater is the density of water (1000 kg/m^3).
The density of the steel is:
[tex]\rho=\frac{\text{mass}}{\text{volume}}[/tex]Based on table 3, you have:
mass = 50.7 g = 0.0507 kg
volume = 0.0000063 m^3
[tex]\rho=\frac{0.0507kg}{0.0000063m^3}\approx8047.62\frac{kg}{m^3}[/tex]Then, for the relative density you obtain:
[tex]\rho_{\text{rel}}=\frac{\rho}{\rho_{\text{water}}}=\frac{8047.62\frac{kg}{m^3}}{1000\frac{kg}{m^3}}\approx8.048[/tex]Hence, the relative density of steel is 8.048
12000 inches to yards
ANSWER
[tex]\begin{equation*} 333.33\text{ yds} \end{equation*}[/tex]EXPLANATION
We want to convert 12000 inches to yards.
To do this, divide the value in inches by 36:
[tex]\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}[/tex]That is the answer.
What does the strength of frictiondepend on?A. The direction of the forces.B. The types of surfaces and how hard theobjects are being pushed.C. The color of surfaces and how hard they push.D. Only how hard the objects are being pushed.
Answer:
B. The types of surfaces and how hard the objects are being pushed.
Explanation:
The force of friction is the result of the
If a rock has 376 J of potential energy when it’s held 10.1 m above the ground what is its mass? Round to the nearest tenth
Answer: 3.8 kg
Explanation:
The formula for calculating potential energy is expressed as
Potential energy = mgh
where
m is the mass of the object
g is the acceleration due to gravity
h is the height above the ground
From the information given,
Potential energy = 376
g = 9.8 m/s^2
h = 10.1
Thus,
376 = m x 9.81 x 10.1
376 = 99.081m
m = 376/99.081
m = 3.8 kg
The mass is 3.8 kg
A person pulls a block 3 m along a horizontal surface by a constant force F=20 N. Determine the work done by force F acting on the block.
Take into account that the work done is given by the following formula:
[tex]W=F\cdot d[/tex]where,
F: applied force = 20N
d: distance = 3m
Replace the previous values into the formula for W:
[tex]W=(20N)(3m)=60J[/tex]Hence, the work done by the person is 60J
Help me with number 1 I’m very lost. I just need the equation. No explanation I’m stuck on 1 part.
Given data:
* The value of angular velocity is,
[tex]\omega=1.3\text{ rad/s}[/tex]Solution:
(a). The time period of the oscillation in terms of the angular velocity is,
[tex]T=\frac{2\pi}{\omega}[/tex]Substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi}{1.3} \\ T=4.83\text{ s} \end{gathered}[/tex]Thus, the time period of oscillation is 4.83 s.
(b). The frequency of the oscillation in terms of the time period is,
[tex]undefined[/tex]A radioactive tracer has a half-life of two hours how much of a 2500 g sample will be available after 18 hours?
ANSWER:
4.88 grams
STEP-BY-STEP EXPLANATION:
We must first calculate how many half-life there are in 18 hours, knowing that each half-life takes 2 hours.
[tex]\frac{18}{2}=9\text{ half-life}[/tex]Now, knowing this, we can calculate the number of grams applying 9 times the half-life, like this:
[tex]\frac{2500}{2^9}=4.88\text{ g}[/tex]Which means that after 18 hours there are 4.88 grams
# 3. If a wire is connected to row 1, column a , and another to row 1, column c, are the two wires connected ?
In a breadboard, internal connections are made between groups of five holes in each row. That means that holes A, B, C, D, and E in row 1 are all connected, but for example, F and A are not connected.
Since 1A and 1C are in the same row and in the same five-hole group, they are connected.
Answer: The two wires are connected.
A 10 gram ball is rolling at 3 m/s. Calculate its kinetic energy.
ANSWER:
0.045 joules
STEP-BY-STEP EXPLANATION:
Given:
mass (m) = 10 g = 0.01 kg
velocity (v) = 3 m/s
The kinetic energy is given by the following formula:
[tex]K_E=\frac{1}{2}mv^2[/tex]We replacing:
[tex]\begin{gathered} K_E=\frac{1}{2}\cdot0.01\cdot3^2 \\ K_E=0.045\text{ J} \end{gathered}[/tex]The kinetic energy is 0.045 joules.
Can anyone help me in this question, Please?
The net force acting on the mass [tex]m_{1}[/tex] is 12 N and the tension (T) in the string is 42 N.
Both the blocks will move with the same acceleration, that is, a = 3 [tex]m/s^{2}[/tex]
Now, from the free-body diagram of the block [tex]m_{1}[/tex]
T - F = [tex]m_{1} a[/tex]
It is given that F = 30 N
and [tex]m_{1} = 4 kg[/tex] , a = 3[tex]m/s^{2}[/tex]
Putting all these values, we get:
T - 30 = 4*3 = 12
T = 12 + 30
T = 42 N
Hence, the tension in the string = 42 N
Now, the net force acting on the mass [tex]m_{1}[/tex] is
[tex]F_{net}[/tex] = T - F = 42 - 30 = 12 N
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How much heat must be removed from 750 grams of water at 0°C to form ice at 0°C?
ANSWER:
250500 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 750 grams
We can determine the amount of heat removed by taking into account the following image:
This means that to go from water to ice there is an absorbed radiation dose of 334 J/g, therefore, the heat removed is:
[tex]\begin{gathered} Q=m\cdot a \\ \\ \text{ We replacing:} \\ \\ Q=750\cdot334 \\ \\ Q=250500\text{ J} \end{gathered}[/tex]The heat removed is 250500 joules.
If you have a convex lens whose focal length is 10.0 cm, where would you place an object in order to produce an image that is virtual?
When an object is placed between first focus and optical center of a convex lens then virtual image is produced.
Here , focal length is 10.0 cm . So object distance should be less than 10.0 cm
Final answer is : between focus and optical center of the lens
This question is based on Oscillations and waves. I tried it for days and I just couldn't get it right.
ANSWER:
The maximun velocity is 16.07 m/s
At x = 0.26
The velocity is 8.36 m/s
The accelearion is 286.67 m/s^2
The resorting force is 86 N
STEP-BY-STEP EXPLANATION:
Given:
k = 310 N / m
Max distance = 0.5 m
Mass of block = 0.3 kg
Max velocity:
Using conservation of energy:
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v^2=\frac{kx^2}{m} \\ \text{ replacing} \\ v^2=\frac{310\cdot0.5^2}{0.3} \\ v=\sqrt[]{258.33} \\ v=16.07\text{ m/s} \end{gathered}[/tex]At x = 0.26 m:
[tex]\begin{gathered} v^2=\frac{kx^2}{m} \\ v^2=\frac{310\cdot0.26^2}{0.3} \\ v=\sqrt[]{69.85} \\ v=8.36\text{ m/s} \end{gathered}[/tex]Acceleration:
[tex]\begin{gathered} F=k\cdot x \\ F=m\cdot a \\ \text{ therefore} \\ m\cdot a=k\cdot x \\ a=\frac{k\cdot x}{m} \\ \text{ replacing} \\ a=\frac{310\cdot0.26}{0.3} \\ a=286.67\text{ }\frac{m}{s^2} \end{gathered}[/tex]The resorting force:
[tex]\begin{gathered} F=m\cdot a \\ \text{ replacing} \\ F=0.3\cdot286.67 \\ F=86\text{ N} \end{gathered}[/tex]For an object starting from rest and accelerating with constantacceleration, distance traveled is proportional to the square of thetime. If an object travels 2.0 furlongs in the first 2.0 s, how far willit travel in the first 4.0 s?
Since the object is accelerating with constant acceleration we can use the following formula for the position of the object:
[tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]where x0 is the initial position, v0 is the initial velocity, a is the acceleration and t is the time. In this case, the initial position and velocity are zero. Plugging the values given we have:
[tex]\begin{gathered} 2=\frac{1}{2}a(2)^2 \\ 2=2a \\ a=1 \end{gathered}[/tex]Hence, the acceleration of the object is 1 furlong per second per second.
Once we know the acceleration we can use the same formula to determine how far the object will travel in four seconds.
[tex]\begin{gathered} x=\frac{1}{2}(1)(4)^2 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Therefore, the object will travel 8 furlongs in four seconds.
The mass number of an atom is found by1) adding the number of protons and neutrons.2) adding the number of protons and electrons.3) subtracting the number of protons from the number of neutrons.4) subtracting the number of protons from the number of electrons.
The mass number of an atom is the sum of number of protons present in the nucleus and number of neutrons present in the same nucleus.
[tex]M=\text{Protons}+\text{Neutrons}[/tex]Thus, 1st option is the correct answer.
A truck can travel at 100 km/hr How long would it take to drive 900km?
Given:
The speed of the truck is,
[tex]v=100\text{ km/hr}[/tex]The distance is,
[tex]s=900\text{ km}[/tex]The time to drive this distance is,
[tex]t=\frac{s}{d}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\frac{900}{100} \\ =9\text{ hrs} \end{gathered}[/tex]Hence, the time is 9 hrs.
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer: ___________ J (round to the nearest whole number)
According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:
[tex]W=\Delta K[/tex]Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:
[tex]K=\frac{1}{2}mv^2[/tex]Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:
[tex]\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\ \\ \therefore K\approx532,346J \end{gathered}[/tex]Therefore, the answer is: 532,346 J.
What is the resistance of an electric frying pan that draws 12 amperes of current when connected to 120 Volt circuit
10 ohms
Explanation
the resistance of an electric circuit is given by:
[tex]\begin{gathered} R=\frac{V}{I} \\ where\text{ R is the resitance} \\ Vi\text{s the voltage} \\ I\text{ is the current} \end{gathered}[/tex]so
Step 1
a) let
[tex]\begin{gathered} R=R \\ V=120\text{ volts} \\ I=12\text{ Amperes} \end{gathered}[/tex]b) now, replace in the expression
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{120\text{ volt}}{12\text{ Amp}} \\ R=10\text{ ohms} \end{gathered}[/tex]therefore, the answer is
10 ohms
I hope this helps you
Water flows through a pipe diameter of 8.000 cm at 49.0 m/min. Find the flow rate in m3/min
We are asked to determine the volumetric flow rate through a pipe of diameter 8.000 cm. To do that we will use the following formula:
[tex]R=Av[/tex]Where:
[tex]\begin{gathered} R=\text{ volumetric flow rate} \\ A=\text{ cross-area of the pipe} \\ v=\text{ velocity of the flow} \end{gathered}[/tex]The cross-area of the pipe is the area of a circle and is given by:
[tex]A=\frac{\pi D^2}{4}[/tex]Where:
[tex]\begin{gathered} A=\text{ cross-area} \\ D=\text{ diameter} \end{gathered}[/tex]Before we determine the area we will convert the diameter from cm to meters using the following conversion factor:
[tex]100cm=1m[/tex]Multiplying by the conversion factor we get:
[tex]8.000cm\times\frac{1m}{100cm}=0.080m[/tex]Now, we plug in the value in the formula for the area:
[tex]A=\frac{\pi(0.080m)^2}{4}[/tex]Solving the operations:
[tex]A=0.005m^2[/tex]Now, we plug in the values of area and velocity in the formula or the volumetric flow rate:
[tex]R=(0.005m^2)(49.0\frac{m}{\min })[/tex]Solving the operations:
[tex]R=0.246\frac{m^3}{min}[/tex]Therefore, the flow rate is 0.246 cubic meters per minute.
A human heart found to beat seventy five times in a minute. Calculate the beat frequency?
[tex]{ \green{ \tt{f = \frac{number \: of \: beats}{time \: taken}}}} [/tex]
[tex]{ \green{ \tt{number \: of \: beats = 75}}}[/tex]
[tex]{ \green{ \tt{time \: taken =1 \: min \: = 60 \: sec}}}[/tex]
[tex]{ \red{ \sf{f = \frac{ \cancel{75^{3}}}{ \cancel{ 60_{4} }}}}}[/tex]
[tex]{ \blue{ \boxed{ \purple{ \sf{f = \frac{3}{4} = 1.2 {s}^{ - 1}}}}}} [/tex]
___________________________________
[tex]{ \blue{ \sf{T = \frac{1}{f}}}} [/tex]
[tex]{ \blue{ \sf{T = \frac{1}{ \purple{ \sf{1.2}}}}}} [/tex]
[tex]{ \boxed{ \red{ \sf{T = 0.8 \: S}}}}[/tex]
A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30°angle to the horizontal. The force of kinetic friction exactly balances the component ofgravity down the plane so that the plate, once started, glides down at constant velocity. If90% of the mechanical energy of the system is absorbed by the aluminum, what is itstemperature increase at the bottom of the incline? (Specific heat for aluminum is 900J/kg⋅°C.) Why do I multiply 15 by sin30?
A scheme of a the given situation is shown below:
First, consider that the work over the plate is done only by the component of the weight parallel to the incline (due to the perpendicular component is balanced by the friction force), then, the work on the plate is:
W = m*g*d*sinθ
where,
m: mass = 0.2kg
d: length of the incline = 15m
g: gravitational acceleration constant = 9.8m/s^2
θ = 30
By replacing the previous values into the expression for W, you obtain:
W = (0.2 kg)(9.8 m/s^2)(15 m)sin(30)
W = 14.7 J
Now, take into account that the amount of heat absorbed by the aluminum plate is given by the following formula:
Q = m*c*ΔT
Q: heat
m: mass
c: specific heat
ΔT: change in tempetaure
Take into account that the 90% of the mechanical energy is absorbed by the plate, which means that 0.9 of the work is converted to absorbed heat by the plate.
Then, you can write:
0.9W = Q
0.9(14.7J) = Q
13.23J = Q
Replace the given expression for Q into the previous equation and solve for ΔT, as follow:
m*c*ΔT = 13.23 J
ΔT = 13.23J/(m*c)
Now, replace the values of m and c for aluminum and simplify:
ΔT = 13.23J/(0.2kg*900J/kg°C)
ΔT = 0.0735°C
Hence, the temperature increase at the bottom of the incline is approximately 0.07°C