Answer:
The speed of a cyclist is 0.3 km/min.
Explanation:
Given
The distance d = 6km Time t = 20 minutesTo determine
We need to determine the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
We know the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speedd = distance coveredt = time takensubstitute d = 6, and t = 20 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{6}{20}[/tex]
Cancel the common factor 2
[tex]s=\frac{3}{10}[/tex]
[tex]s=0.3[/tex] km/min
Thus, the speed of a cyclist is 0.3 km/min.
To increase the gravitational force between the two objects above, I could
Your answer:
A. increase the mass of the objects and decrease the distance between the objects.
B. increase BOTH the distance and the mass between the two objects
C. decrease the mass of the objects and increase the distance between the objects.
D. decrease BOTH the mass and distance between the two objects.
Answer:
Option A
Explanation:
alex often draws his dream house
Answer:
hopefully alex quackity hahhaa
Explanation:
i hope this was free points and not an actual thing
Answer:
cool, cool for alex .....
QUESTION 9 / 10
What is the first step you should take when you want to open a savings account?
A. Present your photo ID to the bank representative,
B. Make your initial deposit.
C. Review the different savings account options that your
bank offers.
D. Go to the bank and fill out an application.
Answer:
A
Explanation:
AWNSER:
awnser:
C
explanation:
a 2kg bowling bowl is at a top of the hill with a speed of 9.8 m/s with a height of 40m. what is its potential energy
To calculate potential energy, use the formula
[tex]ep \: = mgh[/tex]
Where m is mass in kg, g is gravitational field strength in m s^-2 and h is height in metres.
So in this case, calculate
2kg × 9.8m/s × 40m
and you should find your answer.
help please due today
Answer:
equal and opposite
Explanation:
..........
it is an organic compound and an essential micronutrient that the body needs in small amounts.
Answer:
Nutrients the body needs in relatively small amounts are called micronutrients. They include vitamins and minerals. Vitamins are organic compounds that are needed by the body to function properly. ... Vitamins and minerals do not provide energy, but they are still essential for good health
Explanation:
On Earth, the number flux of solar neutrinos from the p-p chain is:
f_neutrino = 2fo/2.62MeV
Other nuclear reactions in the Sun supplement this neutrino flux with a small additional flux of higher-energy neutrinos. A neutrino detector in Japan, named SuperKamiokande, consists of a tank of 50 kton of water, surrounded by photomultiplier tubes. The tubes detect the flash of Cerenkov radiation emitted by a recoiling electron when a high-energy neutrino scatters on it.
Required:
a. How many electrons are there in the water of the detector?
b. Calculate the detection rate for neutrino scattering, in events per day.
Answer:
Explanation:
The volume of the tank = 50 kton
50 kton = 5 × 10⁷ kg
Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³
Then;
5× 10⁷ kg will contain [tex]( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}[/tex]
= 1.67 × 10³⁴ electrons
(b)
Suppose:
[tex]f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}[/tex]
Then;
10⁻⁶ of [tex]f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}[/tex]
[tex]=6.7 \times 10^{4}\ s^{-1} cm^{-2}[/tex]
Thus, the number of high energy neutrinos which will interact with water is:
= [tex]6.7 \times 10^4 \times \sigma[/tex]
= [tex]6.7 \times 10^4 \times 10^{-43}[/tex]
= [tex]6.7 \times 10^{-39} s^{-1}[/tex]
For 1.67 × 10³⁴ electrons, the detection rate is:
[tex]6.7 \times 10^{-39} \times 1.67 \times 10^{34}[/tex]
[tex]= 11.19 \times 10^{-5} \ s^{-1}[/tex]
= 9.668 per day
attraction is seen between the poles of two bar magnet in the case of
Answer:
he magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the centerExplanation:
Help plz with both I’ll mark brainliest
A 150.0 g baseball has a velocity of 28.5 m/s. What is its kinetic energy in J
Answer:
KE = 60.91875 J
Explanation:
First, convert the mass of the ball into kg, since we want the answer in J (SI system):
150 g = 0.15 kg
then use the kinetic energy formula
[tex]KE=\frac{1}{2} m*v^{2} \\KE=\frac{1}{2} (0.15)*(28.5)^{2}\\KE=60.91875 J[/tex]
Which of the following statements is true?
A. Friction primarily affects objects that contain iron.
B. Friction pulls objects toward the center of the Earth
C.
Friction does not affect objects in motion.
D.
Friction slows down or stops objects in motion.
Answer:
D. Friction slow down or stop objects in motion.
What is the difference between the reflection and refraction of light
Answer:
Reflection can simply be defined as the reflection of light when it strikes the medium on a plane. Refraction can be defined as the process of the shift of light when it passes through a medium leading to the bending of light. The light entering the medium returns to the same direction.
Answer:
reflection is your image and refraction is light
3. Consider a large windmill 30m in diameter. On a windy day, suppose that the windmill entrains a stream of air at a speed of 40 mph. Downstream of the windmill, the entrained stream exits over a large diameter at a speed of 20 mph. The pressure is 2atm at the inlet and equals atmospheric pressure at the outlet. Find the power (in megawatts) generated by the windmill. Density of air is 1.2 kg/m3
Answer:
The power generated by the windmill is approximately 1.364 MW
Explanation:
The diameter of the windmill, d = 30 m
The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s
The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s
The pressure at the inlet, P₁ = 2 atm
The pressure at the outlet, P₂ = 1 atm
The density of air, ρ = 1.2 kg/m³
The power obtained from the windmill, 'P', is given as follows;
[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]
Where;
[tex]g_c[/tex] = 1.0 kg/(N·s²)
A = Cross-sectional rea of the the windmill = π·D²/4 = π×(30 m)²/4 = 706.858347 m²
Plugging in the values, we get;
[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]
The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
For the questions below, include units if applicable. If necessary, use a separate sheet of paper for 1, 6c and 7c. Tire pressure is in part a function of the temperature of the tire.
1. Based on everyday experience, state (in words) the relationship between tire pressure and temperature. Look at the data below and see if the numbers support your statement.
2. Prepare a hand-drawn plot of the two variables on the reverse side of this worksheet. Include a title, axis labels (with units), and a trendline. Estimate the tire pressure when the temperature is 18.6°C: Estimate the temperature of the air in the tire when the pressure is 37.0 psi: 3.
a. Prepare a plot using graphing software. Include a title, axis labels (with units), the equation of the best-fit Line and the R? value on the graph.
b. Re-write the equation of the best-fit line substituting "Temperature" for x and "Pressure" for y directly on the graph.
c. Attach the fully labeled graph to this worksheet.
4. What is the value of the slope for the relationship between temperature and pressure?
5. Determine the percent error using the definition of percent error: Use 0.145 psi/" for the "Actual" value of the slope. %error = Actual-Experimental % Error Actual
6. Based on your computer-generated graph,
a. visually estimate the tire pressure when the temperature is 18.6°C:
b. calculate the tire pressure at this temperature using the equation of the best fit line: the graph to ensure that this value is reasonable.
c. compare the calculated pressure to the two visually interpolated values (Steps 2 and 6a). Comment on any discrepancies.
7. Based on your computer-generated graph,
a. visually estimate the temperature of the air in the tire when the pressure is 37.0 psi:
b. calculate the temperature of the air in the tire at this pressure: Use the graph to ensure that this value is reasonable.
c. compare the calculated temperature to the two visually interpolated values (Steps 2 and 7a). Comment on any discrepancies.
Data:
Temperature (x) Tire Pressure, psi (y)
12.9°C 3.39 x 10
15.4C 34.25
-2.10 F 2.68 x 10
19.5 °C 3.50 x 10
29.6 'F 36.53
Answer:
All answer are explained below in the explanation section.
Explanation:
1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.
As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.
Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.
2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.
Tire pressure at temperature 18.6 degree C is ~ 35 psi.
Temperature at air pressure of 37 psi is ~26.1 degree C
3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.
3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32
3. c) It is already done in part a of this question.
4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.
5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]
Plugging in the values, we get:
Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:
% Error = 21.33%
6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi
6. b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.
6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.
7. a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.
7. b.) Estimation of temperature from best fit value of line is = 26.64 degree C
7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.
What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?
Answer:
86.33m/s^2
Explanation:
Acceleration = Force/Mass
= 25.9/0.3
= 86.33
Four-wheel drive trucks do not stop better on icy
roads than a car. Is what law of motion (Newton's laws)
Can someone help please
Answer:
The average blood speed is 0.39 m/s
Explanation:
According to Poiseuille’s law, the volume flow rate, Q (m³/s) of a fluid of
viscosity η through a tube or pipe of radius r and length L is:
Q = πr⁴ΔP/8ηL
where ΔP is the change in pressure or pressure difference of the fluid; L is length of the pipe
Also, the volume flow rate, Q, is related to the average velocity v, by the formula:
Q = Av
where A is the cross-sectional area of the pipe; A = πr²
Q = πr²v
Thus, the Poiseuille’s becomes: πr²v = πr⁴ΔP/8ηL
v = r²ΔP/8ηL
From the given values:
r = 2.5 mm = 0.0025 m
L = 15 cm = 0.15 m
ΔP = 380 Pa
η = 5 * 10⁻³ Pa.s
substituting the given values in the equation
v = {(0.0025)² * 380} / (8 * 5 * 10⁻³ * 0.15)
v = 0.39 m/s
Therefore, the average blood speed is 0.39 m/s
Which of the following will be attracted toward a positively charged cloth?
Positively charged sock
Negatively charged pipe
Sound waves
Light energy
Postive and negatives attract, positive and positive repel. answer is negatively charged pipe.
sound waves and light energy are not "affected" by static electricity
Which 5 are examples of locomotor movements? Select all that apply.
Question 1 options:
Hopping
Galloping
Sitting
Sleeping
Walking
Running
Skipping
Eating
Answer:
Five examples of locomotor movements are:
Hopping
Galloping
Walking
Running
Skipping
Explanation:
I hope it helps ❤❤
Which part of the water cycle is where vapor from plants leaves the plants as they breath?
condensation
Transpiration
evaporation
Answer:
I think it is transpiration
Answer:
transpiration is the right answer
A motorcycle and rider have a total mass equal to 300 kg. The rider applies the brakes, causing the motorcycle to decelerate at a rate of -5 m/s^2. What is the net force on the motorcycle?
Answer:
Net force = - 1500 N
Explanation:
We calculate the net force acting using Newton's second Law:
[tex]F_{net}=m*a\\F_{net}=(300 \,kg)*(-5\,m/s^2)\\F_{net}=-1500\,N[/tex]
For a given substance, the molecules
move fastest when the substance is
Answer:GAS
Explanation:
Which statement accurately describes a relationship between parts of the
universe?
O A. There are millions of nebulae in a star.
O B. There are thousands of universes in a galaxy.
C. There are trillions of galaxies in the universe.
O D. There are billions of galaxies in a solar system.
Answer:
C. There are trillions of galaxies in the universe.
Explanation:
A. is wrong as nebulae are found inside galaxies and inside the universe, not inside stars.
B. is wrong because there are trillions of galaxies in the universe, not the latter.
D. The solar system consists of the eight planets, the Sun, comets, meteors, dwarf planets, and is inside the Milky Way galaxy and thus cannot have galaxies inside it.
Please give Brainliest
Answer:
C.
Explanation:
There are trillions of galaxies in the universe.
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?
Answer:
Explanation:
During an energy transfer, the collision loss for an electron can be determined by using the formula:
[tex]Q = \dfrac{4mME }{(m+M)^2}[/tex]
However; from the total stopping power & power loss of the electron;
[tex]\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}[/tex]
where;
Z = atomic no. for lead = 82
E = 1.9 MeV
∴
radiational energy loss = collisional energy loss [tex]=\dfrac{82 \times 1.9}{800}[/tex]
= 0.19475
b)
Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.
Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.
Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.
Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.
Which example is correctly matched with its type of friction?
A. Pushing a car that isn't moving is an example of slkiding friction.
B. A plane flying through the air is an example of static friction.
OC.
A skateboard wheels on cement is an example of rolling friction.
OD.
A sled sliding down a grassy hill is an example of fluid friction.
Answer:
A. pushing a car that isn't moving
An astronaut named Sandra Bullock has drifted too far away from her spaceshuttle while attempting to repair the Hubble Space telescope. She realizes that theshuttle is moving away from her at 3 m/s. On her back is a 10 kg jetpack which consistsof an 8 kg holding tank filled with 2 kg of pressurized gas. Without the jetpack, sheand her space suit have a mass of 80 kg.
Required:
a. She is able to use the gas to propel herself in the same direction as the shuttle. The gas exits the tank at a uniform rate with a constant velocity of 100 m/s, relative to the tank (and her). After the gas in the tank has been released, what is her velocity?
b. After this, she throws her empty tank into space and relies on the conservation of momentum to increase her speed to match that of the shuttle. With what velocity (in her frame of reference!) will she have to throw the tank?
Answer:
a) v_f = 0.898 m / s, b) v₂ = -6.286 m / s
Explanation:
a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s
Initial instant. Before you start to pass gas
p₀ = (M + Δm) v
M is the mass of the astronaut M = 80Kg and Δm the masses of the gases
Final moment. When you expel the gases
p_f = M (v + Δv) + Δm (v-v_e)
where v_e is the gas velocity v_e = 100 m / s
momentum is conserved
p₀ = p_f
M v + Δm v = Mv + M Δv + Δm v -Δm ve
0 = M Δv - Δm v_e
if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM
M dv = -v_e dM
∫ dv = - v_e ∫ dM / M
We solve, between the lower limits v₀ = v with M = M₀ and the upper limit v = v_f for M = M_f
v_f - v₀ = - v_e (ln M_f - Ln M₀)
v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])
v_f = v₀ + v_e ln (\frac{M_o}{M_f})
let's calculate
v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)
v_f = -1.3 +2.20
v_f = 0.898 m / s
b) launch the jetpack to increase its speed up to the speed of the platform
initial instant. Before launching the tanks
p₀ = (M + m') v_f
final instnte. After launching the tanks
p_f = M v₁ + m' v₂
indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it
p₀ = p_f
(M+ m) v_f = M v₁ + m v₂2
v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]
v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f
let's calculate
v₂ = 80/10 (0.898 - 0) + 0.898
v₂ = -7.1874 + 0.898
v₂ = -6.286 m / s
Please help I’m almost done with exam
Which phrase is the best description of what a telescope does?
O A. Causes objects to grow larger
B. Transports equipment to space
C. Converts solar energy to electricity
D. Detects electromagnethwaves
Answer:
D
Explanation:
Telescopes detect electromagnetic waves from space and it travels back to the telescope lens in order for you to see.
Answer:
D
Explanation:
Ap ex science exam lol
6th grade science I mark as brainliest
Answer:
8. organelle
Explanation:
9. Epithelial tissue
am i correct?
A radio transmitting station operating at a frequency of 125 MHz has two identical antennas that radiate in phase. Antenna B is 9.05 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?
Answer:
Explanation:
The formula for frequency is:
[tex]f = \dfrac{c}{\lambda}[/tex]
If we make [tex]\lambda[/tex] the subject; we have:
[tex]\lambda = \dfrac{c}{f}[/tex]
[tex]\lambda = \dfrac{3\times 10^6 \ m/s}{125 \ MHz (\dfrac{10^6 \ Hz}{1 \ MHz})}[/tex]
[tex]\lambda = 2.4 \ m[/tex]
Let assume that there is a point P between antenna A to B.
where;
A to B = 9.05
A to P = x and
P to B = 9.05 - x
Then, the condition for the constructive inteference is:
Δx = nλ
x - (9.05 - x) = nλ
2x - 9.05 = n(2.4)
So, we need to start assigning values to n so that the value of x becomes less than or equal to 9.05 m
If n = -1
Then;
2x - 9.05 = (-1)(2.4)
x = 3.325 m
If n = -2
Then;
2x - 9.05 = (-2)(2.4)
x = 2.125 m
If n = -3
Then;
2x - 9.05 = (-3)(2.4)
x = 0.925 m
If n = 0
Then;
2x - 9.05 = (0)(2.4)
x = 4.525 m
If n = 1
2x - 9.05 = (1)(2.4)
x = 5.725 m
If n = 2
Then;
2x - 9.05 = (2)(2.4)
x = 6.925 m
Hence, there exist 7 points in which constructive interference occurs.
