Answer:
b
Explanation:
Which of the following best
describes amplitude?
A. Amplitude is how fast a wave travels.
B. Amplitude is how far a wave moves from its resting
position
C. Amplitude is the resting position of a wave.
Answer:
its b
Explanation:
Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.
Question 2 of 15
When the source of a sound is moving, its speed increases.
A. True
B. False
our battery has died and your friends push your vehicle so you can kick-start the engine. You and the vehicle have a combined mass of 1600 kg. If your friends do 6000 J of work and one-third of that is dissipated by friction, how fast is your vehicle traveling?
Answer:2.23 m/s
Explanation:
Given
Mass of person and vehicle is [tex]m=1600\ kg[/tex]
Total work done is [tex]W_t=6000\ J[/tex]
Friction consumes one-third of the energy
The remaining two-third is consumed to increase the kinetic energy
[tex]\Rightarrow \dfrac{2}{3}\times 6000=\dfrac{1}{2}\times 1600\times v^2\\\\\Rightarrow 4000\times 2=1600\times v^2\\\\\Rightarrow v=\sqrt{5}\ \approx 2.23\ m/s[/tex]
Which of the following is NOT true about a space-based internet system? * Data is transmitted in the form of electromagnetic radiation. Signals travel faster through space than through fiber-optic cables. No ground equipment is needed to access the internet. The network is made of many satellites organized in a grid pattern
Answer:
A Data is transmitted in the form of electromagnetic radiation.
B Signals travel faster through space than through fiber-optic cables.
C No ground equipment is needed to access the internet.
D The network is made of many satellites organized in a grid pattern.
Explanation:
THose are the options
Answer:
B
Explanation:
elophase II is the final and the fourth stage in meiosis II when the chromosomes reach the opposite poles of the
Answer: The correct answer is formation of four haploid nuclei.
The event occurring during telophase II includes formation of four haploid daughter cells. Telophase II is the final and the fourth stage in meiosis II when the chromosomes reach the opposite poles of the nuclear spindle. During this stage both the daughter cells get divided forming four haploid cells. The development of the nuclear envelope around each set of the chromosome and cytokinesis also takes place during telophase II.
Explanation:
???
the equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z
Answer:
x = 1, y = 1 and z = 0
Explanation:
Given equation;
[tex]P^x V^y T^z = constant[/tex]
Boyle's law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
Mathematically the law is written as;
[tex]PV = constant[/tex]
From the given equation, the values of x, y and z that will match this law is calculated as follows;
[tex]P^1 V^1 T^0 = constant\\\\P^1 = P\\\\V^1 = V\\\\T^0 = 1\\\\P \times V \times 1 = PV = constant\\\\Thus, x = 1, \ y = 1 \ \ and \ z = 0[/tex]
Which is an example of a wedge?
A.
B
C
D
Answer:
B I believe, because the axe is wedged into the log
Answer:
B
Explanation:
I think just rhis is the answer
A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closest point. (a) Calculate the semimajor axis of the orbit. Incorrect: Your answer is incorrect. m (b) Calculate the eccentricity of the orbit. Incorrect: Your answer is incorrect. Did you find the semimajor axis a from the greatest and smallest radii
Answer:
a) 6636 km
b) 0.0154
Explanation:
The height above the earth at its furthest point is 368 km
The height above the earth at its closest point is 164 km
Radius of the Earth is 6370 km
The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km
The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km
If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.
Basically,
2a = R + r
a = (R + r) / 2
a = (6738 + 6534) / 2
a = 13272 / 2
a = 6636 km
Eccentricity, e = (a - r) / a
Eccentricity, e = (6636 - 6534) / 6636
Eccentricity, e = 102 / 6636
Eccentricity, e = 0.0154
HELP ITS DUE IN 4 MINUTES
Answer:
the nephew
Explanation:
because we need to First find the trails that is on top of all the others in order to find who was the last to leave, since the person that was last to leave would most likely step on another person's trail that has been theirs before them.
by looking at this picture I noticed that the butler's footprint was the latest one but the dogs footprint steps over it therefore making it the last to leave, and the person who owns a dog is her nephew so therefore the nephew is the one that stole the butterfly trophy.
An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right, and proceeds an additional 4.0 m, reaching the far corner of the same wall from which it began. If the second leg of the journey was crawled in half the amount of time as the first, what was the ant's average speed for the whole trip?
Answer:
vavg = 0.37 m/s
Explanation:
The average speed is just the relationship between the total distance traveled, and the total time required for that travel , as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} (1)[/tex]
We know that for the first leg of the journey, the ant crawls at a constant speed of 0.24 m/s, moving 3.0 m.We can find the time required for this part, just applying the definition of average velocity, and solving for the time t (which we will call t₁), as follows:[tex]t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)[/tex]
From the givens, we know that the time for the second part is exactly the half of the value found in (2), so we can write the total time Δt as follows:[tex]\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)[/tex]
We also know that in the second leg of the journey, the ant traveled 4.0 m, which adds to the 3.0 m of the first part, making a total distance of 7.0 m.Per definition of average speed, we can write the following expression as in (1) replacing Δx and Δt by their values, as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)[/tex]
Conductivity in aqueous solutions, is a measure of the ability of water to conduct an electric current.
a. True
b. False
Answer:
true
Explanation:
i think its true beacuse water can cause electricity
pls let me know if corrct
if yes pls mark brainlest
An astronomer studies an image taken by a satellite in space and sees an area of gas, dust, and many stars distributed along spiral arms. What is the astronomer most likely seeing?
Answer:the answer is galaxy
Explanation:
16. Two capacitors have an equivalent
capacitance of 30 pF, if connected in
parallel, and 7.2 pF, if connected in
series. Find C1 and C2.
Answer:
C1 + C2 = 30 parallel connection
C1 * C2 / (C1 + C2) = 7.2 series connection
C1 * C2 = 7.2 * (C1 + C2) = 216
C2 + 216 / C2 = 30 using first equation
C2^2 + 216 = 30 C2
C2^2 - 30 C2 + 216 = 0
C2 = 12 or 18 solving the quadratic
Then C1 = 18 or 12
HELP ITS DUE IN 4 MINUTES
Answer:
The nephew
Explanation:
The nephew has a seeing eye dog which means the nephews tracks are the dogs tracks, and the dog tracks are not covered by anything which means the dog was last and stole the butterfly.
Hope this helps!
A marshmallow is fired from ground level with an initial speed of 41.5 m/s at an
angle of 33.5° above the horizontal. (a) Determine the maximum height reached
by the marshmallow. (b) Determine the horizontal range that the marshmallow
travels during its flight.
Calculate the average maximum height for all three trials when the mass of the bottle is 0.125 kg, 0.250 kg,
0.375 kg, and 0.500 kg.
Record your calculations in Table A of your Student Guide.
When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is
m.
When the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is
m.
When the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is
m.
When the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is
m.
Answer:
10
Explanation:
i dont know this is right answer
0.35
0.91
1.26
1.57
Explanation:
I did it on ed2020 and got it correct
Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, and immediately jumps back up into the air. Right before she hits the force plate, her vertical velocity is 3.0 m/s downwards. After leaving the ground again, her vertical velocity is 3.5 m/s upwards. Katelyn was in contact with the ground for 0.4 seconds. (a) What was the impulse exerted on Katelyn when she was on the force plate
Answer:
J = 357.5 kg*m/s
Explanation:
The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:[tex]p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)[/tex]
By the same token, the final momentum is as follows:[tex]p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)[/tex]
As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):[tex]J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)[/tex]
The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 75-kg (165-lb) person of height 1.83 m (6 ft) has a body surface area of approximately 2.0 m2. (a) What is the net amount of heat this person could radiate per second into a room at 18°C (about 65°F) if his skin’s surface temperature is 30°C? (At such temperatures, nearly all the heat is infrared radiation, for which the body’s emissivity is 1.0, regardless of the amount of pigment.) (b) Normally, 80% of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person’s basal metabolic rate.
Answer:
(a) Q = 142.67 W
(b) Basal Metabolic Rate = 178.33 W
Explanation:
(a)
We can find the heat radiated by the person by using Stefan-Boltzman's law:
[tex]Q = \sigma A (T^4 - T_{s}^4)\\[/tex]
where,
Q = heat radiated per second = ?
σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴
A = Surface Area = 2 m²
T = Temperature of Skin = 30° C + 273 = 303 k
Ts = Temperature of room = 18° C +273 = 291 k
Therefore,
[tex]Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4][/tex]
Q = 142.67 W
(b)
Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,
[tex]Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}[/tex]
Basal Metabolic Rate = 178.33 W
(a) The excess heat just through radiation Q = 142.67 W
(b) Basal Metabolic Rate = 178.33 W
What is metabolic rate?Basal metabolic rate is the number of calories your body needs to accomplish its most basic (basal) life-sustaining functions.
(a) We can find the heat radiated by the person by using Stefan-Boltzman's law:
[tex]Q=\sigma A(T^4-T_s^4)[/tex]
where,
Q = heat radiated per second = ?
σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴
A = Surface Area = 2 m²
T = Temperature of Skin = 30° C + 273 = 303 k
Ts = Temperature of room = 18° C +273 = 291 k
Therefore,
[tex]Q=(5.6703\times 10^{-8}\times (2)\times[(303)-(291)][/tex]
Q = 142.67 W
(b) Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,
[tex]\rm Q=0.8\times (Basal\ Metabolic\ Rate)[/tex]
[tex]\rm Basal\ metabolic \ Rate = \dfrac{Q}{0.8}[/tex]
[tex]\rm Basal\ Metabolic\ Rate = \dfrac{142.67}{0.8} =178.330W[/tex]
Basal Metabolic Rate = 178.33 W
Hence the excess heat just through radiation Q = 142.67 and Basal Metabolic Rate = 178.33 W
To know more about Basal metabolic rate Follow
https://brainly.com/question/14095049
1. An electron travels 4.82 meters in 0.00360 seconds. What is its average speed?
Answer:
speed =distance /time
speed =4.82/0.00360
speed =1338.8m/s
A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion. x(t)
Answer:
x(t) = -8sin2t
Explanation:
See the attachment for solution
From my solving, we can deduce that w² = 4, and thus, w = 2
Therefore, the general solution is
x(t) = c1 cos2t + c2 sin2t
Considering the final variable, we can conclude that
x(0) = 0
x'(0) = -8 m/s
The final solution, thus
x(t) = -8sin2t
A 700 kg horse has 4000 J of kinetic energy as it is running
What is the velocity of the horse?
Answer:
m
Explanation:
While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude of the change in momentum of the ball?
Answer:
5.94 N·s
Explanation:
F = 132 N
t = 0.045 s
Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s
We would like to compare the angular momentum of Mars about its axis of rotation with that of Earth's. The mass of Mars is 11% that of Earth, with a radius 53% that of Earth, and a rotational period 103% that of Earth. Assuming both planets to be uniform spheres calculate the ratio of the angular momentum of Mars to that of Earth.
Explanation:
Assume that mass of Earth is M, radius of earth orbit is R, and rotational period of Earth is T.
The angular momentum of Earth is,
[tex]L_{z} &=M R^{2} \omega \\ &=M R^{2}\left(\frac{2 \pi}{T}\right) \\ &=\frac{2 \pi M R^{2}}{T}[/tex]
The mass of mars is, mass of Earth
=0.11 M
The radius of mars orbit is, of radius of earth
=0.53 R
The rotational period of mars is, of period of Earth
=1.03 T
The angular momentum of mars is,
[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]
The angular momentum of mars is,
[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]
The ratio of angular momentum of mars to that of earth is,
[tex]\frac{L_{m}}{L_{E}}=\frac{\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}}{\frac{2 \pi M R^{2}}{T}} \\ \frac{L_{m}}{L_{E}}=0.03 \\ \frac{L_{m}}{L_{B}}=3.0 \times 10^{-2}[/tex]
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the adjacent Figure. It collides with an identical disk that is at rest before the collision. The collision is elastic. Disk 1 goes off at an 60 5.0 m/s angle of 60 with respect to the x-direction. Disk 2 g 30 goes off at an angle of 30 with respect to the x-direction. What best describes the speeds of the disks after the collision?
Solution :
Let [tex]$m_1=m_2=4$[/tex] kg
[tex]$u_1 = 5$[/tex] m/s
Let [tex]$v_1$[/tex] and [tex]$v_2$[/tex] are the speeds of the disk [tex]$m_1$[/tex] and [tex]$m_2$[/tex] after the collision.
So applying conservation of momentum in the y-direction,
[tex]$0=m_1 .v_1_y -m_2 .v_2_y $[/tex]
[tex]$v_1_y = v_2_y$[/tex]
[tex]$v_1 . \sin 60=v_2. \sin 30$[/tex]
[tex]$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$[/tex]
[tex]$v_2=1.732 \times v_1$[/tex]
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
[tex]$m_1.u_1=m_1.v_1_x+m_2.v_2_x$[/tex]
[tex]$u_1=v_1_x+v_2_x$[/tex]
[tex]$5=v_1. \cos 60 + v_2 . \cos 30$[/tex]
[tex]$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$[/tex]
[tex]$v_1 = 2.50$[/tex] m/s
So, [tex]$v_2 = 1.732 \times 2.5$[/tex]
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
Which of the following statements is NOT correct about sea breezes?
A) In a sea breeze, the sand is much warmer than the water, therefore creating a low pressure above the sand.
B) The breeze/wind moves from the sea to the land.
C) The breeze/wind moves from the land to the sea.
D) There is high pressure above the sea, and low pressure above the sand.
The breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.
What is sea breeze?Local wind patterns known as sea breezes flow from the sea to land during the day. When there is no strong large-scale wind system and it is very hot or very cold during the day or at night, sea breezes and land breezes alternate along the coasts of large lakes or oceans.
Because the sea breeze's surface flow ends over land, an area of low-level air convergence is created. Locally, this convergence frequently causes air to rise, which promotes the formation of clouds. Showers over land in the afternoon may result from such clouds.
Hence, the sea breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.
Learn more about sea breeze here:
https://brainly.com/question/13015619
#SPJ6
1. Describe the components of the reflex arc
Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 5.0 A .
How close can a credit card be held to this wire without damaging its magnetic strip?
Answer:
his distance is too small (r = 0.01 mm), therefore the cut can be at any distance
Explanation:
For this exercise let's use the ampere law.
Let's use a cylinder as the circulating surface
∫ B. ds = μ₀ I
in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1
B 2π r = μ₀ I
r = [tex]\frac{\mu_o I}{2\pi B}[/tex]
The field needed to demagnetize the card is B = 1000 gauss = 0.1 T
r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]
r = 2 10⁻⁷ 5.0/0.1
r = 1 10⁻⁵ m
this distance is too small (r = 0.01 mm), therefore the cut can be at any distance
an ice skater is moving across a flat and level skating rink and is speeding up. which one of the following statement is true of the ice skater
a) its potential energy is constant
b) its potential energy is increasing
c) its potential energy is decreasing
Answer:
A
Explanation:
The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:
d
→
L
d
t
=
0
or
→
L
=
→
l
1
+
→
l
2
+
⋯
+
→
l
N
=
constant
.
Note that the total angular momentum
→
L
is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.
Which sentence accurately uses the homophones “they’re,” “there,” or “their”?
Many of the students left there backpacks on the bus.
They’re going to come home as soon as the movie is over.
I think I left the bags of groceries on the floor over their.
These dogs bark at everyone, but there not dangerous at all.
Answer:
They're going to come home as soon as the movie is over.
Answer:
B: They're going to come home as soon as the movie is over.
Explanation:
Pick 3 safety rules in science that you think are important and explain what would happen if the rules were no followed. Used your own word.
Answer:
Answer down below!
Explanation:
Always wear goggles! This is to prevent chemicals or any other substance to get into your eyes. If this wasn't followed, more people could get hurt.Clean up afterwards! Its always better to clean up so someone else couldn't get hurt! If people don't follow this, it could be a mess and someone could get seriously injured.When smelling something, waft it to your face, and not get real close. This is to prevent getting chemicals into your body that could be poisonous. If people didn't folloe this, someone could be seriously sick.