Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
Answer:
Final velocity = 119.83 m/s
Final elevation = 731.9 m
Explanation:
We are told that temperature of air changes from 0 to 10°C
Thus;
Change in temperature; ΔT = 10 - 0 = 10°C
Also, its velocity changes from zero to a final velocity. Thus;
v1 = 0 m/s
v2 is unknown
Also, its elevation changes from zero to a final elevation.
So, z1 = 0 and z2 is unknown
Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.
Thus Equating the formula for both kinetic and internal energy gives;
½m(v2² - v1²) = mc_v•ΔT
m will cancel out and v1 is zero to give;
v2² = 2c_v•ΔT
v2 = √(2c_v•ΔT)
Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K
Thus;
v2 = √(2 × 718 × 10)
v2 = √14360
v2 = 119.83 m/s
To find z2, we will equate potential energy formula to that of the internal energy.
Thus;
mg(z2 - z1) = mc_v•ΔT
m will cancel out and since z1 is zero, then we have;
z2 = (c_v•ΔT)/g
z2 = 718 × 10/9.81
z2 = 731.9 m
Answer:
Explanation:
The rate of conductive heat transfer in watts is:
q = (k/s) A ΔT
where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.
a)
Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:
q/A = (52.4 / 0.014) (210)
q/A = 786,000 W/m²
b)
Given that A = 0.42 m², we can find q:
q = (0.42 m²) (786,000 W/m²)
q = 330,120 W
A watt is a Joule per second. Convert to Joules per hour:
q = 330,120 J/s * 3600 s/hr
q = 1.19×10⁹ J/hr
c)
If we change k to 1.8 W/m/K:
q = (k/s) A ΔT
q = (1.8 / 0.014) (0.42) (210)
q = 11,340 J/s
q = 4.08×10⁷ J/hr
d)
If k is 52.4 W/m/K and s is 0.024 m:
q = (k/s) A ΔT
q = (52.4 / 0.024) (0.42) (210)
q = 192,570 J/s
q = 6.93×10⁸ J/hr
The planet that Punch should travel to in order to weigh 118 lb is Pentune.
<h3 /><h3 /><h3>The given parameters:</h3>
- Weight of Punch on Earth = 236 lb
- Desired weight = 118 lb
The mass of Punch will be constant in every planet;
The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;
where;
- M is the mass of Earth = 5.972 x 10²⁴ kg
- R is the Radius of Earth = 6,371 km
For Planet Tehar;
For planet Loput:
For planet Cremury:
For Planet Suven:
For Planet Pentune;
For Planet Rams;
The weight Punch on Each Planet at a constant mass is calculated as follows;
Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.
<u>The </u><u>complete question</u><u> is below</u>:
Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.
Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).
<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>
Learn more about effect of gravity on weight here: brainly.com/question/3908593
Answer:
The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.
The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.
Explanation:
Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.
Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.
