Answers
Answer:
It cannot be constant because if it does not change and each time it increases its strength and speed.
Explanation:
Related Questions
An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)
Answers
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
Helppp!!! It’s due right now. Review question.
Answers
Answer:
A person that consistently runs 3 meters evert second
Explanation:
Because as a human walks 3 meters every 1 second it isn't consider uniform. It has to be in a erect motion
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:__________.
Answers
Complete question is;
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:
garage door openers: 40.0 MHz
standard cordless phones: 40.0 to 50.0 MHz
baby monitors: 49.0 MHz
FM radio stations: 88.0 to 108 MHz
cell phones: 800 to 900 MHz
Global Positioning System: 1227 to 1575 MHz
microwave ovens: 2450 MHz
wireless internet technology: 2.4 to 2.6 GHz
Which of the following statements correctly describe the various applications listed above? Check all that apply.
a.) All these technologies use radio waves, including low-frequency microwaves.
b.) All these technologies use radio waves, including high-frequency microwaves.
c.) All these technologies use a combination of infrared waves and high-frequency microwaves.
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.
Answer:
B, D, E, F are the correct statements.
Explanation:
Looking at the options;
A) This is true because radio waves are electromagnetic radiation being used today in television, mobile phones, radios and other areas of communication technologies. And the examples given to us fall in the category of technologies that use radio waves.
B) microwaves usually have long wavelengths and low frequencies. However, sometimes they could have high frequencies usually more than radio waves. Thus, this option is correct.
C) This option is wrong because it's not all the listed technologies that use combination of infrared waves and high-frequency microwaves.
D) we are given the frequency of microwave ovens as 2450 MHz.
Converting to GHz gives; 2.45 GHz.
We are told that wireless internet technology has frequency between 2.4 to 2.6 GHz. Thus, microwave frequency falls in the same range as wireless internet technology and thus the statement is true.
E) we know that wavelength is inversely proportional to frequency. This means that the higher the frequency, the shorter the wavelength.
In the frequencies given to us, wireless internet technology have the highest frequency which means they have the shortest wavelength. The statement is true.
F) from the frequencies given to us, the smallest is garage door openers = 40.0 MHz = 40 × 10^(6) Hz while the biggest is 2.6 GHz = 2.6 × 10^(9) Hz
Formula for wavelength is;
Wavelength = speed of light/frequency
Speed of light = 3 × 10^(8) m/s
Thus;
Wavelength = (3 × 10^(8))/(40 × 10^(6))
Or wavelength = (3 × 10^(8))/(2.6 × 10^(9))
So,wavelength = 7.5 m or 0.12 m
This falls into the given range of 0.10 to 10.0 m.
Thus, the statement is true.
What are the five classes of objects that orbit the sun?
Answers
The following statements address the science behind the pulley system illustrated:
A. The pulleys increase the entropy of the system.
B. The force applied to the rope is less than the force needed to lift the object.
C. The pulleys help generate as much energy as possible.
D. The pulleys multiply energy input, resulting in more energy output.
E. The pulleys generate no thermal energy.
Which of these statements is/are true?
i. Statements A and B
ii. Statements D and E
iii. Only statement C
iv. All of the statements
Answers
Answer:
i. Statements A and B
Explanation:
Sana nakatulong
An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?
Answers
Answer:
500km per hour
Explanation:
if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answers
Answer:
nzkdjdksishdjsdjjdjnzkskejeoueeuieeoyrie ryrhdhcksodopdncndnszdkhfoeosheiwuef wokxkzdkjdoeehxjbxbdkeiehdhdhddjjddjdhhdhdhhhjdhfjdjjfjddhdheudiwiehdbejwowud
Explanation:
isos
Which of the following best describes our
atmosphere?
A. envelope of gases that surround Earth
B. a specific range of altitude where plant life flourishes
C. The air, water, and land that form our planet
D. the water vapor in the air surrounding our planet
Answers
The energy goes from _____ to the _____ above it.
Answers
Answer:
The energy goes from the ground state to the excited states above it.
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answers
Answer:
B i think
Explanation:
...
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
Answers
To increase the potential energy of the system, what did you have to do?
Answers
Answer:
You can use work to add kinetic energy to a system or to increase potential energy in the system.
Explanation:
Potential energy stored in any system can be released as kinetic energy. Kinetic energy can be transformed to do work or to increase potential energy.
hope this helped
A dentist uses a concave mirror to examine a tooth that is 1.00
cm in front of the mirror. The image of the tooth forms 10.0 cm
behind the mirror. What is the mirror's radius of curvature?
Answers
Answer:
-2.2cm
Explanation:
A dentist uses a concave mirror to examine a tooth that is 1.00 cm in front of the mirror. The image of the tooth forms 10.0 cm behind the mirror. What is the mirror's radius of curvature 0.909 m.
To determine the mirror's radius of curvature, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror
v is the image distance (distance of the image from the mirror)
u is the object distance (distance of the object from the mirror)
In this case, the object distance (u) is 1.00 cm and the image distance (v) is -10.0 cm (since the image is formed behind the mirror, the distance is negative).
1/f = 1/(-10.0) - 1/(1.00)
Simplifying the equation:
1/f = -0.1 - 1
1/f = -1.1
Now, we can find the reciprocal of both sides of the equation:
f/1 = -1/1.1
f ≈ -0.909 m
The negative sign indicates that the mirror is concave.
Therefore, the mirror's radius of curvature is approximately 0.909 m.
To know more about Mirror :
https://brainly.com/question/116014
#SPJ2.
What kind of energy is in a moving skateboard
Answers
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answers
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
Which of the following is NOT something
the atmosphere does?
A. traps in warmth
B. keeps Earth in a spherical shape
C. provides oxygen to breathe
D. protects Earth from meteoroids
Answers
Answer:
I think the answer is B, keeps Earth in a spherical shape
What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?
Answers
Answer:
[tex]f=4.70\times 10^{14}\ Hz[/tex]
Explanation:
Given that,
The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]
We need to find the frequency of the light. We know that,
[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]
So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].
Light rays enter a transparent material. Which description best describes what happens to the light rays
Answers
When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.
A 0.70-kg disk with a rotational inertia given by MR 2/2 is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs from the free end. If the string does not slip then as the mass falls and the cylinder rotates the suspension holding the cylinder pulls up on the mass with a force of______
Answers
Answer:
The force will be "9.8 N".
Explanation:
The given values are:
mass,
m = 0.7 kg
M = 2
g = 9.8
Now,
⇒ [tex]\tau = T \alpha[/tex]
then,
⇒ [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]
⇒ [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]
⇒ [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]
⇒ [tex]=8.34 \ m/s[/tex]
hence,
⇒ [tex]T=mg+M(g-a_t)[/tex]
On substituting the values, we get
⇒ [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]
⇒ [tex]=6.86+2(1.46)[/tex]
⇒ [tex]=6.86+2.92[/tex]
⇒ [tex]=9.8 \ N[/tex]
Please respond to this for 15 points. Please don’t put in a link.
Answers
Answer:
e. Combustion
Explanation:
In Combustion reaction, a substance reacts with oxygen from the air and resultant product is that it releases carbon dioxide and water.
Here,
2C2H6 is the substance that reacted with 7O2 (Oxygen) to release 4CO2 (Carbon Dioxide) and 6H2O (Water).
The weight of a column of air pushing
down over an area is called which of
Help Resources
these?
A. density
B. volume
C. mass
D. air
pressure

Answers
Answer:
air pressure
:::::::::::::::::
air pushes the weight of a column
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answers
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
Help me please with both questions?
Answers
Answer:
question #1 is A
Question #2 is C
Explanation:
Which of the following best describes wind?
A А
Sinking warm air moving a few feet above the ground
B
An air current formed by changes in ocean tides
с
Cool air rushing in to fill an area of low pressure
D
Rising warm air pushing cool air down toward Earth
Answers
The correct statement about the wind is:
Cool air rushing in to fill an area of low pressureWhat is the wind?Wind is the movement of air currents in relation to the Earth's surface, which is caused by pressure differences and air movement.
Characteristics of the windIt is a meteorological phenomenon originated in the movements of rotation and translation of the Earth.When the rising air cools and loses the moisture it was carrying, due to condensation and rain, the result is dry and cool air.Therefore, we can conclude that the wind is the current of air that occurs in the atmosphere due to natural causes, from high pressure areas to low pressure areas.
Learn more about Characteristics of the wind here: https://brainly.com/question/11463167
Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answers
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
Easy question just don’t understand it please help.
Answers
30/0.75=40
Answer: 40km/h
A go-cart is traveling at 15 mi/hr. How long does it take the go-cart to travel 3 miles?
Answers
Answer:
12 min
Explanation:every 4 minutes is 1 mile
A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent
Answers
Answer:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
[tex]f_o = \frac{V}{2L}[/tex]
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]
Therefore, the wave represents the second harmonic.
L = λ
The potential energy of a mass-spring system when the spring is fully compressed and the mass is at rest is 200 J. After releasing the mass, assuming there is no dissipative force, the system will oscillate. At a point during the oscillation the potential energy of the system is 50 J. What is the kinetic energy of the mass at that point
Answers
Answer:
150 J
Explanation:
If there is not any dissipative force in the system, the mass and the spring will oscillate eternally., but of course, we assume this is a theoretical situation. The conservation of energy in a system implies that the sum of the potential energy plus kinetic energy remains constant, therefore if in the initial point the mass has 200 J (potential energy) and is at rest ( kinetic energy = 0) the overall energy at the beginning is 200 J. At any point of the oscillation if the potential energy is 50 J the kinetic energy must be 150 J.
What is the reason that the moon looks dimmer before eclipse? Why does it take some time to be brighter again after eclipse?
Answers
Answer:
Why does it take sometimes to be brighter again after eclipse? The moon looks dimmer before lunar eclipse because the moon enters into shadow of penumbra region of earth. As a result, the brightness of moon decreases and looks dimmer.
Explanation:
Answer: The moon looks dimmer before lunar eclipse because the moon enters into shadow of penumbra region of earth. As a result, the brightness of moon decreases and looks dimmer.
Explanation: