Answer:
it is cool and warm
Explanation:
Which describes the greatest displacement?
A. walking 3 m east, then 3 m north,
then 3 m west
B. walking 3 m east, then 3 m south,
then 3 m east
C. walking 3 m north, then 3 m south,
then 3 m north
D. walking 3 m north, then 3 m west,
then 3 m south
Tabial 1
A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.277 m/s^2 in the x direction. After 46.6 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find:
a. The magnitude.
b. The direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answer:
a) 25.76 m/s
b) 30°
Explanation:
See attachment
The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.
Required:
a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?
b. How do you know the answer in part (B) to be correct?
Complete Question
The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.
Required:
a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?
b
Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.10 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?
c. How do you know the answer in part (B) to be correct?
Answer:
a
The two possible frequencies of the second propeller in Hz
[tex]f_1 = 11.58\ Hz [/tex]
[tex]f_2 = 7.58 \ Hz [/tex]
The two possible frequencies of the second propeller in rpm
[tex]f_1 = 695 \ rpm [/tex]
[tex]f_2 = 455 \ rpm [/tex]
b
The correct answer for the frequency of the second single-bladed propeller is
[tex]f_1 = 695 \ rpm [/tex]
c
The above answer is correct because when the beat frequency of the second propeller increases(i.e from 2.0 Hz to 2.10 Hz) the frequency of the second propeller becomes much greater than that of the first propeller so looking at the two possible value of frequency of the second propeller (i.e 695 rpm and 455 rpm ) we see that it is 695 rpm that is showing that increase of the second propeller compared to the first propeller
Explanation:
From the question we are told that
The number of engines is n = 2
The number of propellers is m = 2
The angular frequency of the single-bladed propellor [tex]w = 575 rpm[/tex]
The frequency of the beat heard at this velocity is [tex]f = 2.0 \ Hz[/tex]
Converting the beat frequency to rpm
[tex]f = 2 * 60 = 120 \ rpm[/tex]
Generally the the two possible frequencies of the second propeller in rpm is
[tex]f_1 = w + f[/tex]
=> [tex]f_1 = 575 + 120[/tex]
=> [tex]f_1 = 695 \ rpm [/tex]
And
[tex]f_2 = w - f[/tex]
=> [tex]f_2 = 575 - 120[/tex]
=> [tex]f_2 = 455 \ rpm [/tex]
Converting the angular frequency of the single-bladed propellor to rpm
[tex]w = \frac{575}{60} [/tex]
[tex]w = 9.58 \ Hz [/tex]
Generally the the two possible frequencies of the second propeller in
Hz is
[tex]f_1 = w + f[/tex]
=> [tex]f_1 = 9.58 + 2[/tex]
=> [tex]f_1 = 11.58\ Hz [/tex]
And
[tex]f_2 = w - f[/tex]
=> [tex]f_2 = 9.58 - 2[/tex]
=> [tex]f_2 = 7.58 \ Hz [/tex]
What is the displacement of the object?
Answer:
Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Ancient gold coins have been found in Greece and Italy.What property allow for gold to be stamped into coins ?
Answer: Malleability
Explanation:
Gold in addition to being so rare, is the most malleable element there is. A Malleable element can be shaped and reshaped without it breaking and gold is so malleable that you can hit it till it becomes so flat, its transparent.
This malleability allowed for the ancient peoples to be able to shape and stamp gold into coins and their rarity made them quite valuable. Even today, investors still shape gold into exquisite designs of coins so it not only has monetary value, but aesthetic value as well.
The heat flux, q, from a hot body in a stationary fluid is known to be a function of the fluid’s thermal conductivity k, and kinematic viscosity, ν, the temperature difference ΔT , the length of the object L, and the product of the local gravitational constant and the thermal expansion coefficient for the fluid, gβ. Determine the number of Pi groups thatcan be formed from these six parameters.
Answer:
The number of pi groups are P = 3 pi groups
Explanation:
From the question we are told that
The number of parameters is n = 6
These parameters are : Fluid thermal conductivity : k
Kinematic viscosity: v
The temperature difference ΔT
The length of the object L
Gravitational constant g
Thermal expansion coefficient β
The number of the basic dimensional unit is N = 3
Mass M
Length L
Time T
Generally the number of Pi groups is mathematically evaluated as
P = n - N
=> P = 6 - 3
=> P = 3 pi groups
Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind of helmet uses an airbag that deploys from a pouch worn around the rider's neck. In tests, a headform wearing the inflated airbag is dropped from rest onto a rigid platform; the speed just before impact is 6.00 m/s. Upon impact, the bag compresses its full 12.0 thickness, slowing the headform to rest.
Required:
Determine the acceleration of the head during this event, assuming it moves the entire 12.0 cm.
Answer:
acceleration = -15.3g
Explanation:
given data
speed = 6.00 m/s.
thickness = 12
moves the entire = 12.0 cm
solution
we will use here equation that is
v² - u² = 2 × a × s ........................1
here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration
so we put here value and get acceleration
a = [tex]\frac{v^2-u^2}{2s}[/tex]
a = [tex]\frac{0^2-6^2}{2\times 0.12}[/tex]
a = -150 m/s² ( negative sign means it is a deceleration )
and
acceleration in units of g
a = [tex]\frac{-150}{9.8}[/tex]
a = -15.3 g
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volume charge density rho(r) of this thin shell distribution in terms of σ and an appropriate delta function. Verify explicitly that the units of your final expression are correct. Also show that your total integrated charge comes out right.
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴
[tex]\rho (r) \ \alpha \ \delta (r -R)[/tex]
[tex]\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)[/tex]
[tex]\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)[/tex]
To find the constant k, we examine the total charge Q which is:
[tex]Q = \int \rho (r) \ dV = \int \sigma \times dA[/tex]
[tex]Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2[/tex]
∴
[tex]\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2[/tex]
[tex]\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]
[tex](2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]
Thus;
[tex]k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2[/tex]
[tex]k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2[/tex]
[tex]k * R^2= \sigma \times R^2[/tex]
[tex]k = R^2 --- (2)[/tex]
Hence, from equation (1), if k = [tex]\sigma[/tex]
[tex]\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}[/tex]
[tex]\mathbf{\rho (r) =0 \ \ at \ \ r<R \ \ or \ \ r>R}[/tex]
To verify the units:
[tex]\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}[/tex]
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
[tex]Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr[/tex]
[tex]Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma *R^2[/tex] since [tex]( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )[/tex]
[tex]\mathbf{Q = 4 \pi R^2 \sigma }[/tex]
ou step off the limb of a tree clinging to a 30-m-long vine that is attached to another limb at the same height and 30-m distant. Assuming air resistance is negligible, how fast are you gaining speed at the instant the vine makes an angle of 25 degrees with the vertical during your descent
Answer:
v = 7.42 m / s
Explanation:
For this exercise we will use the conservation of mechanical energy.
Starting point. Before jumping from the tree
Em₀ = U = m g h
Final point . Part of the trajectory at 25º
Em_{f} = K + U = ½ m v2 + m g y
as they indicate that there is no air resistance, mechanical energy is conserved
Em₀ = Em_{f}
m g h = ½ m v² + m g y
v² = 2g (h - y)
Let's use trigonometry to find the height that has descended, how the angle is measured with respect to the vertical
cos 25 = y / L
y = L cos 25
we substitute
v² = 2 g (h - L cos 25)
for this case h = L = 30 m
v2 = 2g L (1- cos25)
let's calculate
v² = 2 9.8 30 (1 -cos 25)
v = √55.09
v = 7.42 m / s
Some common types of forces that you will be dealing with include the gravitational force (weight), the force of tension, the force of friction, and the normal force. It is sometimes convenient to classify forces as either contact forces between two objects that are touching or as long-range forces between two objects that are some distance apart. Contact forces include tension, friction, and the normal force. Long-range forces include gravity and electromagnetic forces. Note that such a distinction is useful but not really fundamental: For instance, on a microscopic scale the force of friction is really an electromagnetic force. In this problem, you will identify the types of forces acting on objects in various situations.
Now consider a different situation. A string is attached to a heavy block and is used to pull the block to the right along a rough horizontal table.
a. What is the upward force that acts on the book called?
1-tension
2-normal force
3-weight
4-friction
b. Which object exerts a force on the block that is directed to the right?
1-the block itself
2-the earth
3-the surface of the table
4-the string
Answer:
No. A:
1-tension
No. B:
4-the string
Explanation:
The upward force that acts on the book is called "tension" while the "string" is the object that exerts a force on the block that is directed to the right.
As the string is pulled, the tension exerts an upward force on the block. The frictional force acts on the block to the left. So, both the tension and friction will act on the block in order to effect its pulling on the surface of the table.
According to the graph given, what is the best explanation for what is occurring as energy us removed?
A) an exothermic change in which water freezes
B) an exothermic change in which water boils
C) and endothermic change in which rubbing alcohol evaporates
D) and endothermic change in which water freezes
Answer:
I believe it's A) "An exothermic change in which water freezes I could be wrong though just giving my opinion
Explanation:
___ is the pull that all objects exert on each other.
A.Resistance
B.Inertia
C.Gravity
D. Force
Answer:
B
Explanation: Inertia is where it exerts.
If the strength of the action force is halved, what happens to the strength of the reaction force?
Answer:
it is reduced to half of that strength
Una cubeta de agua puede girar en un círculo vertical sin que el agua se derrame, incluso en lo alto del círculo cuando la cubeta está boca abajo. ¿Por qué sucede esto? ¿Que otros ejemplos puedes mencionar que utilicen el mismo principio?
Answer:
v> √ gr
Explanation:
Let's analyze this problem from Newton's second law
At the lowest point
N - W = m a
at the highest point
W = m a’
in both cases there is a net force towards the center of the circle that we can call the centripetal force, which is responsible for changing the direction and magnitude of the acceleration.
When the bucket is in the highest part, the centripetal force is equal to the weight of the water, but since it carries a horizontal speed, until it starts to fall, it is moving and therefore they follow the bottom of the tube. This implies that there is a minimum speed for this to occur
v> √ gr
This principle is applied in many things, for example the roller coaster, centrifuges, simulators of the effect of acceleration on people.
Think about a hot air balloon travelling around the world in 11 days. How can a balloon travel so far and fast without a engine or other system on board to move the balloon horizontally
Answer:
?
Explanation:
A sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
a. What is the sprinters velocity at the end of the 200 m?
b. How long does it take him to cover it?
c. What is his average velocity?
Answer:
a. 28.28
B.14.4
C.14.4
Explanation:
A. v^2=u^ + 2as
v^2=0^2 + 2*2*200
B. v=u+at
t=v-u/a
C. V+u/2
The sprinter velocity at the end of the 200 m would be 28.28 m/s.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
As given in the problem a sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
By using the third equation of the motion,
v² - u² = 2×a×s
v² - 0² = 2×2×200
v² = 800
v =28.28 m/s
The sprinter velocity at the end of the 200 m would be 28.28 m/s
By using the second equation of motion
S = ut + 1/2*a*t²
u= 0 m/s , a= m/s² s = 200 m
200 = 0 + 0.5*2*t²
t² = 200
t = √200
t = 14.14 seconds
It would take him 14.14 seconds to cover.
average velocity = initial velocity + final velocity /2
= 0+ 28.28/2 m/s
= 14.14 m/s
Thus, The sprinter velocity at the end of the 200 m would be 28.28 m/s.
Learn more about equations of motion from here
brainly.com/question/5955789
#SPJ2
A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels
640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration.
Answer:
(a) The average velocity is 16 m/s
(b) The acceleration is 0.4 m/s^2
(c) The final velocity is 24 m/s
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:
[tex]v_f=v_o+at\qquad\qquad [1][/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]
(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.
We know the distance is x=640 m and the time taken t= 40 s, thus:
[tex]\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16[/tex]
The average velocity is 16 m/s
Using the equation [1] we can solve for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
(c) From [2] we can solve for a:
[tex]\displaystyle a= 2\frac{x-v_ot}{t^2}[/tex]
Since vo=8 m/s, x=640 m, t=40 s:
[tex]\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4[/tex]
The acceleration is 0.4 m/s^2
(b) The final velocity is calculated by [1]:
[tex]v_f=8+0.4\cdot 40[/tex]
[tex]v_f=8+16=24[/tex]
The final velocity is 24 m/s