If a plane covers a distance of 700 miles in 1 hour and 43 minutes, what is its speed? A) Give your answer in mph. B) Give your answer in ft/second
Given:
Distance covered by the plane, d = 700 miles
Time, t = 1 hour and 43 minutes.
Let's find the speed of the plane.
To find the speed, apply the formula:
[tex]\text{Speed = }\frac{dis\tan ce}{time}[/tex]Where:
distance = 700 miles
time = 1 hour 43 minutes
(A) speed in mph.
mph is miles per hour.
Where:
60 minutes = 1 hour
Thus, to find the speed the time is to be in hours.
We have:
[tex]t=1\frac{43\text{ minutes}}{60\text{ minutes}}=1\frac{43}{60}=1.716666\text{ hrs}\approx1.72\text{ hrs}[/tex]Thus, to find the speed in mph, we have:
[tex]\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{700}{1.72}=407.8\text{ mph}[/tex]Therefore, the speed of the plane in mph is 407.8 mph
(B) To find the speed in ft/second
Let's first convert the distance from miles to feet
Where:
1 mile = 5280 feet
700 miles = 700 x 5280 = 3696000 feet
Also convert the time to seconds.
Where:
1 hour = 60 minutes x 60 seconds = 3600 seconds
1.7166 hours = 1.7166 x 3600 = 6180 seconds
Thus, we have:
Distance = 3696000 feet
Time = 6180 seconds
[tex]\text{Speed = }\frac{dis\tan ce}{\text{time}}=\frac{3696000}{6180}=598.06\text{ ft/second}[/tex]Therefore, the speed in ft/second is 598.06 ft/second
ANSWER:
(A) 407.8 mph
(B) 598.06 ft/second
In the calculation of the area of a rectangle with dimensions of 4.282 m by 0.050 m, which of the following answers has the correct number of significant figures?
We have
[tex]A=l\times w[/tex]in our case
l=4.282 m
w=0.050m
we substitute
[tex]A=(4.282)(0.050)=0.2141m^2=0.21m^2[/tex]ANSWER
0.21m^2
A girl weighs 58kg and runs up a staircase 52m high in 23seconds. Find her minimum power output
Given,
The mass of the girl, m=58 kg
The height of the staircase, h=52 m
The time she takes to run up the staircase, t=23 s
The work the girl does to climb up the staircase is equal to the change in the potential energy of the girl.
Thus the work done is given by,
[tex]W=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} W=58\times9.8\times52 \\ =29.56\times10^3\text{ J} \end{gathered}[/tex]The power output is equal to the rate at which the work is done
Thus the power output is given by,
[tex]P=\frac{W}{t}[/tex]On substituting the known values,
[tex]\begin{gathered} P=\frac{29.56\times10^3}{23} \\ =1285.22\text{ W} \end{gathered}[/tex]Thus the maximum power output is 1285.22 W
You call out in a loud voice and hear an echo of your voice. Where are you most likely located?A.In a boat on a lakeB.In an open fieldC.In a canyonD.In a forest
We will have the following:
Since sound bounces off of surfaces particularly ones that are perpendicular, then in this scenario you are most likely located in a canyon.
Identify each setup and describe the difference in the magnetic field created by each.
In A. we have a magnetic field created by a coil in physics, the term coil refers to a long, thin loop of wire, which produces a magnetic field when an electric current is passed through it. in this case, the core inside the coil is the air
In B. We have a magnetic field created by a magnet that is a piece of iron or other magnetic material that has its component with atoms so ordered that the material have properties of magnetism, such as attracting other iron-containing objects or aligning itself in an external magnetic field
In C. Also we have a coil but this time with a core of metal material, a core can increase the magnetic field to thousands of times the strength of the field of the coil alone.
Question 24 (1 point)—A beam of white light hits a prism and is dispersed into the its spectral componentsred, orange, yellow, green, blue, indigo, and violet. If a second prism is placed so thatonly the red band of light can pass through, what will happen?No light will pass through the second prism because red light does not haveenough energyThe prism will disperse the red light into many shades of redThe prism will polarize the red lightThe prism will refract the red light but it will still be the same color of red
prism can disperse a compound light
red is a single coloured light
when it will pass through the prism it will get refracted again but the output will be the red one
so the answer is
The prism will refract the red light but it will still be the same color of red
a cannon sitting on top of a 40 m mound shoots a projectile with an initial velocity of 4.47 m/s parallel to ground. how far away did it land?
Given that,
The initial height of the projectile, y₀=40 m
The initial velocity of the projectile, v₀=4.47 m/s
The direction of the initial velocity is parallel to the ground. Therefore the x-component of the initial velocity v₀x=4.47 m/s.
And the y-component of the initial velocity is v₀y=0 m/s
From the equation of the motion we have,
[tex]y=y_0+v_{0y}t+\frac{1}{2}gt^2[/tex]Where y is the final height of the projectile which is zero as it finally hits the ground. And g =-9.8m/s² is the acceleration due to gravity. And t is the time interval of the flight of the projectile.
On substituting the known values in the above equation,
[tex]\begin{gathered} 0=40+0+\frac{1}{2}\times-9.8\times t^2 \\ =40-4.9t^2 \end{gathered}[/tex]On rearranging the above equation and simplifying it,
[tex]\begin{gathered} t=\sqrt{\frac{40}{4.9}} \\ =2.86\text{ s} \end{gathered}[/tex]The x-component of the velocity remains constant as there is no acceleration in that direction.
The horizontal distance travelled or the range of the projectile can be calculated using the formula,
[tex]R=v_{0x}t[/tex]On substituting the known values in the above equation,
[tex]R=4.47\times2.86=12.78\text{ m}[/tex]Therefore the projectile will land 12.78 meters away
The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 903 Hz?
ANSWER:
0.38 meters.
STEP-BY-STEP EXPLANATION:
Given:
The speed of sound in air (v) = 340 m/s
Frequency (f) = 903 Hz
We calculate the wavelength using the following formula:
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \text{ We replacing:} \\ \\ \lambda=\frac{340}{903} \\ \\ \lambda=0.37652\cong0.38\text{ m} \end{gathered}[/tex]The wavelength is equal to 0.38 meters.
The potential difference across the ends of a wire is doubled in magnitude. If Ohm's law is obeyed, which one of the following statements concerning the resistance of the wire is true?a- The resistance is one half of its original value.b- The resistance decreases by a factor of four.c- The resistance is twice its original value.d- The resistance is not changed.e- The resistance increases by a factor of four.
Given:
The potential difference across the ends of a wire is doubled in magnitude
To find:
which one of the following statements concerning the resistance of the wire is true
Explanation:
The potential difference is doubled in magnitude. We know, according to Ohm's law,
[tex]V=IR[/tex]I is the current and R is the resistance.
If the potential difference increases, the product of I and R also increases.
Now the resistance depends as,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ \rho\text{ is the resistivity, l is the length and A is the area of the conductor.} \end{gathered}[/tex]The resistance is independent of potential differences. When the potential difference increases, the current also increases and the resistance remains the same.
Hence, The resistance is not changed.
A sports car with a mass of 2000 kg travels at a velocity of 20 m/s. What would be true of its momentum if the same sports car was traveling at a velocity of 40 m/s?
The true statement about the sports car is that the momentum of the sports car would be increased.
What is momentum?The term momentum is defined as the product of the mass and the velocity. The momentum of an object is a vector quantity. What it means is that both the magnitude and the direction of the momentum is important.
Now we are told that the sports car has a mass of 2000 kg travels at a velocity of 20 m/s. The momentum at this point is 40000 Kgm/s. If the velocity of the sports car was increased to 40 m/s, the momentum would increase to 80,000 Kgm/s.
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5. Draw the ray diagram to show the image formed by a concave lens with focal length 5cm whenthe object is placed 15 cm from the lens. What is the magnification?
Given:
The focal length of the concave lens is f = -5 cm
The object distance from the lens is u = -15 cm
Required:
Ray diagram
Magnification
Explanation:
The rules related to incident ray and reflected ray are:
1) Any incident ray moving parallel to the principal axis will pass through focus after reflection.
2) Any incident ray passing through the focus will move parallel to the principal axis after reflection.
3) Incident ray passing through the optical center will reflect undeviated.
The ray diagram is shown below
Here, yellow lines represent incident rays passing parallel to the principal axis and the reflected ray passing through the focus.
The maroon line shows the incident ray passing through the optical center and the undeviated reflected ray.
The green color is the object and image.
In order to find magnification, first, we need to calculate the image distance given as
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ =-\frac{1}{5}-\frac{1}{15} \\ v=-3.75\text{ cm} \end{gathered}[/tex]The magnification will be
[tex]\begin{gathered} m\text{ = }\frac{v}{u} \\ =\frac{-3.75}{-15} \\ =0.25\text{ } \end{gathered}[/tex]Final Answer: Thus, the magnification is 0.25 which means the size of the image is 0.25 times the object.
Object A, which has been charged to +13.13 nC, is at the origin. Object B, which has been charged to -26.7 nC, is at x=0 and y=2.72 cm. What is the magnitude of the electric force on object A?
ANSWER:
0.004259 N
STEP-BY-STEP EXPLANATION:
Given:
qA = +13.13 nC
qB = -26.7 nC
d = y - x = 2.72 - 0 = 2.72 cm = 0.0272 m
We use Coulomb's law to calculate the force:
[tex]F=k\cdot\frac{q_A\cdot q_B}{d^2}[/tex]We substitute each value and calculate the force:
[tex]\begin{gathered} F=9\times10^9\cdot\frac{(13.13\times10^{-9})(26.7\times10^{-9})}{(0.0272)^2} \\ \\ F=0.004259\text{ N} \end{gathered}[/tex]The magnitude of the electric force on object A is 0.004259 N
How to calculate electric field on a charge
Let the charge be denoted by q.
Let the total electric force on the charge be denoted by F.
We have to the electric field on a charge.
Electric field on a charge is given by the formula
[tex]E=\text{ }\frac{F}{q}[/tex]An element with an atomic number of 90 and an atomic mass of 230 would have what atomic number after Alpha decay?88909186
Given data:
The atomic number of
A 49 lb piece of steel at 670 F is dropped into 95 lbs of water at 50 F. What is the final temperature of the mixture.
ANSWER
[tex]83.82F[/tex]EXPLANATION
Parameters given:
Mass of steel, M = 49 lb = 22.23 kg
Initial temperature of steel, ts = 670 F = 627.59 K
Mass of water, m = 95 lb = 43.09 kg
Initial temperature of water, tw = 50 F = 283.15 K
We want to find the final temperature of the mixture of the piece of steel and water.
The amount of heat released in the system must be equal to the total amount of heat absorbed by the system:
[tex]Q_R=Q_A[/tex]The piece of steel releases heat while the water absorbs the heat. This implies that the equation above becomes:
[tex]-MC(T_s-t_s)=mc(T_w-t_w)_{}[/tex]where C = specific heat capacity of steel = 468 J/kgK
c = specific heat capacity of water = 4184 J/kgK
Ts = final temperature of steel
Tw = final temperature of water
Since the final temperatures of both the steel and water are equal, it implies that:
[tex]T_s=T_w=T[/tex]Substituting the given values into the equation above, we have that:
[tex]-22.23\cdot468\cdot(T-627.59)=43.09\cdot4184\cdot(T-283.15)[/tex]Simplify the equation above:
[tex]\begin{gathered} -10403.64(T-627.59)=180288.56(T-283.15) \\ \Rightarrow-10403.64T+6529220.428=180288.56T-51048705.76 \\ \Rightarrow180288.56T+10403.64T=6529220.428+51048705.76 \\ \Rightarrow190692.2T=57577926.19 \\ T=\frac{57577926.19}{190692.2} \\ T=301.94K \end{gathered}[/tex]Let us now convert to Fahrenheit:
[tex]T=\frac{(301.94-273.15)\cdot9}{5}+32=83.82F[/tex]That is the final temperature of the mixture.
66) The fundamental wavelength for standing sound waves in an empty pop bottle is 0.88 m. (a) What is the length of the bottle? (b) What is the frequency of the third harmonic of this bottle?
Explanation
Step 1
Let
[tex]\begin{gathered} \text{fundamental wavelength=0.88 ,} \\ \lambda=2L \\ \text{hence, replace} \\ 0.88m=2\text{ L} \\ \text{divide both sides by 2} \\ \frac{0.88m}{2}=\frac{2\text{ L}}{2} \\ 0.44=L \end{gathered}[/tex]hence, the length of the bottle is 0.44 meters
Step 2
What is the frequency of the third harmonic of this bottle?
For the third harmonic, the length of the string is equivalent to three-halves of a wavelength,so
[tex]\begin{gathered} third\text{ armonci= }\frac{\text{3}}{2}\cdot0.88\text{ m} \\ third\text{ armonci= }1.32\text{ m} \\ \end{gathered}[/tex]so
[tex]\begin{gathered} 1.32=2L \\ \end{gathered}[/tex]replace in the quation and solve for
Let speed of sound = 343 m/s
[tex]\begin{gathered} f=\frac{v}{\lambda} \\ \text{replacing} \\ f=\frac{343}{0.88} \\ f=389.77\text{ Hz} \end{gathered}[/tex]I hope this helps
you
What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s?Question 18 (2 points) What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m radi curve at 12 m/s?
2304 Newton is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s
centripetal force=mv^2/r
m=1,600-kg
v= 12 m/s
r= 100-m
centripetal force=mv^2/r
centripetal force=(1,600-kg×12 m/s×12 m/s)÷100-m
centripetal force=230400/100
centripetal force=2304 N
Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. Centripetal, which means "center seeking," is the name of this force.
The Centripetal Force Formula states that the centripetal force equals the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This formula predicts that the centripetal force will double with a doubling of tangential velocity.
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A ball is thrown vertically upward with a speed of 36.0 m/s.(a) How high does it rise? m(b) How long does it take to reach its highest point? s(c) How long does the ball take to hit the ground after it reaches its highest point? s(d) What is its velocity when it returns to the level from which it started? m/s
ANSWER:
(a) 66.1 m
(b) 3.67 s
(c) 3.67 s
(d) -36 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 36 m/s
Final velocity (v) = 0 m/s
(a)
We can calculate the height using the following formula:
[tex]\begin{gathered} h=\frac{v^2-u^2}{2g} \\ \\ \text{ We replacing:} \\ \\ h=\frac{0^2-(36)^2}{(2)(-9.8)} \\ \\ h=66.1\text{ m} \end{gathered}[/tex](b)
Now, we calculate the time as follows:
[tex]\begin{gathered} v=u+gt \\ \\ t=\frac{v-u}{g} \\ \\ \text{ we replacing} \\ \\ t=\frac{0-36}{-9.8} \\ \\ t=3.67\text{ s} \end{gathered}[/tex](c)
The time it takes for the ball to hit the ground after reaching its highest point is the same as the time it takes to reach it, which means that it is equal to 3.67 seconds
(d)
The velocity it takes to return is also the same but being in the opposite direction, it has the same magnitude but with the opposite sign, that is, with a minus sign.
So the velocity when it returns to the level it started from is -36 m/s
URGENT!! ILL GIVE
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Jack is selling lemonade at a stand outside his house. He is charging $1.50 per cup. He needs to make at least enough money to cover the cost of his supplies, which were $15.25.What is the minimum number of cups of lemonade Jack needs to sell?
Given,
Charge of one cup of lemonade, c=$1.50
The amount Jack needs, T=-$15.25
The number of cups Jack needs to sell in order to have the given amount can be calculated as,
[tex]N=\frac{T}{c}[/tex]On substituting the known values,
[tex]\begin{gathered} N=\frac{15.25}{1.5} \\ =10.17 \end{gathered}[/tex]The number of cups of lemonade that can be sold should be a whole number. If Jack sells 10 cups, which is lesser than the answer, he will have less amount than he wanted.
Thus he has to sell 11 cups.
Therefore the minimum number of cups Jack needs to sell is 11.
Answer: I have all answers in screen shots
Hope this helps
If you have a concave mirror whose focal length is 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object, where should you place the object?
given
[tex]\begin{gathered} focal\text{ length=100 cm} \\ let\text{ object distance is u form the pole of mirror } \\ so\text{ u=-u} \\ and\text{ v=?} \\ here\text{ v is the distance of image from the pole of mirror } \\ let\text{ height of the object h} \\ so\text{ h=+h} \\ according\text{ to the question image of the object is } \\ \text{ h}^{^{\prime}}=+10h \\ \end{gathered}[/tex]since the image is in the upright direction so the image is virtual.
and in the case of the mirror, an image can be virtual if and only if
the image is placed between the pole and the focus. so the object
must be placed between the pole and the focus.
[tex]\begin{gathered} the\text{ formula for magnification is given by;} \\ m=-\frac{v}{u}=\text{ }\frac{h^{^{\prime}}}{h} \\ and\text{ the mirror formula is given by the following} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \end{gathered}[/tex][tex]\begin{gathered} -\frac{v}{u}=\frac{h^{^{\prime}}}{h} \\ by\text{ putting all the values} \\ -\frac{v}{u}=\frac{10h}{h} \\ v=-10u \end{gathered}[/tex][tex]undefined[/tex]The Round Up carnival ride below has a radius of 3.90 meters and rotates 0.527 times per second. As shown, riders can be held up by only friction. What coefficient of friction is needed to keep the riders from sliding down? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The radius of the carnival ride is r = 3.9 m.
* The linear frequency of oscillation is f = 0.527 Hz.
Solution:
The angular frequency (angular speed) of the oscillation is,
[tex]\begin{gathered} \omega=2\pi f \\ \omega=2\pi\times0.527 \\ \omega=3.31\text{ rad/s} \end{gathered}[/tex]The frictional force acting on the ride is,
[tex]F=\mu mg[/tex]The force acting on the ride in terms of the angular speed is,
[tex]F=m\omega^2r[/tex]The frictional force acting on the ride is equal to the force acting on the ride for circular motion,
[tex]\begin{gathered} \mu mg=m\omega^2r \\ \mu g=\omega^2r \\ \mu=\frac{\omega^2r}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} \mu=\frac{(3.31)^2\times3.9}{9.81} \\ \mu=4.36 \end{gathered}[/tex]Thus, the coefficient of friction is 4.36.
Which formula is equivalent to
v = m / D
v = mD / v
v = D x m
v = D / m
Answer:
i don't know.
Explanation: I = me Don't = Don't know = know
A basketball player hit the basket eighteentimes during the game. He scored a totalof 30 points, two points for each field goaland one point for each free throw. If hethrew 6 more field goals than free throws,how many field goals did he make? Howmany free throws did he make?
Answer:
fxn g
Explanation:
ggnhbdfhbz
A car starts from rest and accelerates uniformly for a distance of 137 m over an 9.6-second time interval. The car's acceleration ism/s².
In order to solve this question, we will need to use kinematics
Let's see what is given to us:
Distance traveled is 137 meters
Time elapsed is 9.6 seconds
and inital velocity is 0 m/s
Since we are trying to find acceleration, we can use this formula
[tex]\Delta x=v_0t+\frac{1}{2}at^2[/tex]Where Δx is the distance traveled, v0 is the inital velocity, t is time, and a is acceleration
Plugging in what we have, we get
137 = 0(9.6) + 1/2(a)(9.6)^2
Solving for a, we get 2.97 m/s^2
The volume of a cylinder with height h and radius r can be found using the formula V = πr2hFind the volume of a cylinder with radius 8 feet and height 7 feet.__ ft3
In order to calculate the volume of this cylinder, let's use the given values of radius and height in the formula.
Using π = 3.1416, we have:
[tex]\begin{gathered} V=\pi r^2h \\ V=3.1416\cdot8^2\cdot7 \\ V=1.407.44 \end{gathered}[/tex]So the volume is 1407.44 ft³.
Hi I just answered a question for electricity grade 12 could you please if it is correct
F = k*(q1)*(q2)/r^2
We know that:
F = 71 N
k = coulomb's constant = 9*10^9
q1 = ?
q2 = ?
We know that q1 = q2, since in the question it states that the charges are identical.
r = 0.99 m
F = k*(q1)*(q1)/r^2
F = k*(q1)^2/r^2
71 = 9*10^9 * (q1)^2 / 0.99^2
Isolating (q1)^2,
71 * 0.99^2 / (9*10^9) = (q1)^2
Taking square root of both sides to get q1 to the power of 1,
sqrt( 71 * 0.99^2 / (9*10^9) ) = q1
Now, simplifying the left side in a calculator, we get that
q1 = 8.793 * 10^-5 C = q2
It is hard to see the motion of stars because of their extreme distance from us.
It is hard to see the motion of stars because of their extreme distance from us is a True statement.
The locations of the stars in the sky remain largely constant with respect to one another. Taurus is constantly close to Orion. Draco is continually around Ursa Minor. The stars are a continuous backdrop painting when we observe the sky; they don't move much in relation to one another.
While we orbit the sun throughout the year, there are some minute movements. Stars farther away move slightly more left and right than stars closer to us. Even if it has a minor effect, it makes sense. You'll notice that distant items move less in relation to you than nearer ones if you stare out into the distance while you go along the street.
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Is science physics? If so I have a question on it
Inner Planets:
A planet whose orbit lies within the asteroid belt is called inner planets. Mercury, Venus, Earth and Mars are called th inner planets.
Outer Planets:
A planet whose orbit lies outside the asteroid belt is called outer planets. Jupiter, Saturn, Uranus and Neptune are called Outer Planets.
Two 4.137 cm by 4.137 cm plates that form a parallel-plate capacitor are charged to +/- 0.531 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 2.198 mm?
The electric field of two parallel plates is given by:
[tex]\begin{gathered} E\cdot d=\frac{Q}{d\cdot A\epsilon o} \\ so: \\ E=\frac{Q}{A\epsilon o} \end{gathered}[/tex]Where:
[tex]\begin{gathered} \epsilon o=8.8542\times10^{-12} \\ A=0.04137^2=1.71\times10^{-3}m^2 \\ Q=0.531\times10^{-9}C \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} E=\frac{0.531\times10^{-9}}{(8.8542\times10^{-12})(1.71\times10^{-3})} \\ E\approx35071V/m \end{gathered}[/tex]