Answers
Answer:
truuuuuuuuuuuuueeeeeeeee
Related Questions
Using complete sentences and your own words, describe an example of the water cycle in action. Please use at least two of our key terms for today (Water Cycle, Evaporation, Condensation, Precipitation, Crystallization, Run-off, Transpiration, Aquifer) can you please help me i will give brainlyest it says collage but it is middle school pleas help me.
Answers
Answer:
Where does the water that forms rain come from? How about fog? Where does that come from? Have you ever been asked these questions by people, especially those kids who keep asking, 'Why?' Has someone ever told you that the water falling as snow has always been here, or that the water we use was once dinosaur blood, or that we are drinking someone's sweat, or, worse yet, drinking someone else's… gulp? How can this be possibl…
Explanation:
How long does it take a 1.51 × 104 W steam engine to do 8.72 × 106 J of work? Round your answer to three significant figures
Answers
A 4.5-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
Answers
Answer:
The electrical energy is 13.5 kWh or 48600 kJ
Explanation:
Given that,
The power of the resistance heater, P = 4.5 kW
The water heater runs for 3 hours to raise the water temperature to the desired level.
We need to find the amount of electric energy used in both kWh and kJ.
Energy = Power × time
E = 4.5 kW × 3 h
= 13.5 kWh
Since,
1 kWh = 3600 kJ
13.5 kWh = 48600 kJ
Hence, the required electrical energy is 13.5 kWh or 48600 kJ
Which ray diagram demonstrates the phenomenon of absorption?
tum
Answers
Answer:
hjbyvtf ghbj
Explanation:
uhgbnm,likjh
PLEASE HELP THIS IS TIMED!
a constant force is causing a sled to accelerate at 4 m/s^2. If the mass is suddenly cut in half what will the new acceleration be
Answers
How long will it take a planet 150 million kilometers from a star
of mass 1.99 x 1030
kg to complete one full orbit?
Answers
Answer: 370 days.
Explanation:
It will take the planet 16.69 minutes to complete one full orbit.
Data Given;
r = 150*10^6kmmass of star = 1.99*10^30kgGravitational ForceApplying gravitational and centripetal force
[tex]F = \frac{Gm_1m_2}{r^2}\\ F = \frac{mv^2}{r}\\ \frac{mv^2}{r}=\frac{Gm_1M2}{r^2}\\ v = \frac{2\pi r}{T} \\ \frac{m(2\pi r/T)^2}{r} = \frac{GM_1M_2}{r^2}\\ \frac{4\pi ^2r^2}{T^2} = \frac{GM}{r} \\ T^2 = \frac{4\pi^2 r^3}{Gm}\\ [/tex]
Let's substitute the values into the equation
[tex]T^2 = \frac{4\pi ^2 * (150*10^6)^3}{6.67*10^-11 * 1.99 *10^30}\\ T^2 = 1002799.605 \\ T = 1001.398s\\ T = \frac{1001.398}{60} = 16.69min [/tex]
It will take the planet 16.69 minutes to complete one full orbit.
Learn more on planetary rotation here;
https://brainly.com/question/21222010
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Answers
Answer:
6755727gvbu7euyeue77377365353663636
Explanation:
ghfjfkjfjfn has a long way is a good relationship is a different person than the difference is that the answer is 8AM is a good relationship with the other 4AM in a certain way is a different type and have different effects and feelings and the difference is the difference is that you can do it for a long period and the same way you know what to smome but you can get to the claire family a lot of times and the difference between the two is that the person you want and you know it would have different opinions on the subject and then use the difference between the two and a half hour and the difference is that the answer is yes but if you're not going anywhere in the past two and your life you can do that with your 6 or an 6th year old girl who is a good things go to
What is the pharmacy technician’s role in medication therapy management?
Answers
Medical and pharmacy record gathering, organization, and assessment Medication reconciliation (developing a Personal Medication List) Correspondence with patients, physicians, other health care providers. Follow-up to enhance patient adherence
HELPPPP if this object is accelerating to the right at 2.5 m/s2, arrow C
7.0 N
65 N
27 N
23 N
Answers
On a horizontal surface is located
an object of mass 20 kg and it is observed that
To set it in motion, a
80 N force, whereas if the object is
in motion, only one
40 N force, to keep it at speed
constant. Determine the values of the coefficients
static and kinetic friction, respectively.
(g = 10 m / s2).
A) 0.32; 0.11
B) 0.36; 0.15
C) 0.40; 0.2
D) 0.45; 0.22
Answers
By Newton's second law, the net vertical force acting on the object is 0, so that
n - w = 0
where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that
80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is
40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: µ and ν are the Greek letters mu and nu)
The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. Given the oil drop to be 1.00 um in radius and have a density of 920 kg/m^3.
a. Find the weight of the drop.
b. If the drop has a single excess electron, find the electric field strength needed to balance its weight.
Answers
Answer:
a) [tex]W=3.78\cdot 10^{-14}\: N[/tex]
b) [tex]E=2.36\cdot 10^{5} N/C[/tex]
Explanation:
a) The weight is just the mass of the drop times the acceleration of gravity.
[tex]W=mg[/tex]
Or in terms of density.
[tex]W=\rho Vg[/tex] (1)
If we consider the volume of a drop as spherical, we can find the volume (R=1 μm = 0.000001 m).
[tex]V=\frac{4}{3}\pi R^{3}[/tex]
[tex]V=\frac{4}{3}\pi (10^{-6})^{3}[/tex]
[tex]V=\frac{4}{3}\pi 10^{-18}[/tex]
[tex]V=4.19\cdot 10^{-18}\: m^{3}[/tex]
Then, using equation (1), the weight will be:
[tex]W=920\cdot 4.19\cdot 10^{-18}\cdot 9.81[/tex]
[tex]W=3.78\cdot 10^{-14}\: N[/tex]
b) To balance the weight, the electric field times the charge must be the same value in the opposite direction, which means:
[tex]W-qE=0[/tex]
[tex]E=\frac{W}{q_{electron}}[/tex]
[tex]E=\frac{3.78\cdot 10^{-14}}{1.60\cdot 10^{-19}}[/tex]
[tex]E=2.36\cdot 10^{5} N/C[/tex]
I hope it helps you!
The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of the horizontal velocity (which is constant) and the amount of time the pumpkin is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 14 m/s, and fire the pumpkin several times while varying the angle between the cannon and the horizontal. For which angle is the range a maximum (with the initial speed held constant)
Answers
Answer:
x(max) = 20 m
Explanation:
In projectile shoot e have:
V₀y = V₀*sin θ
V₀ₓ = V₀*cos θ
V₀ initial speed
θ shotting angle
Vₓ = V₀ₓ = V₀*cos θ during the trajectory
Vy = V₀y - g*t
And when Vy = 0 h is maximun and the time to reach y maximum is half of the overall time.
According to that
Vy = 0 V₀y = g*t t = V₀y/g
t = 14* sinθ / 9,8
t = 1,43 *sinθ s
And overall time is T = 2* 1,43*sinθ
t = 2,86*sinθ
x = V₀ₓ * t s
x = 14*cosθ * 2,86* sinθ
x = 40 * cosθ * sinθ
Lets take two very well know sin cos pairs
for ∡ 30 ∡ 45 ∡60
and 30 and 60 are complementary angles for boths
cosθ * sinθ is the same ( 1/2)*(√3/2)
cosθ * sinθ = 0,433 and
x = 40* 0,433
x = 17,32 m
For ∡ 45 boths sin45 and cos45 are equal √2/2
In this case
x = 40*(√2/2)*(√2/2)
x = 40* 2/4
x = 20 m
And that is the maximum range
x(max) = 20 m
Which is an example of current electricity?
Answers
Answer quick please thank
Answers
If the evaporator outlet temperature on an r410A system is 50f and the evaporator superheat is 10f, what is the evaporating pressure of the refrigerant in the system
Answers
Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
Two forces whose resultant is 100newton are perpendicular to each other. If one of them makes an angle of 60newton with the resultant. Calculate it magnitude
Answers
Answer:
100 N
Explanation:
Given that,
Two forces whose resultant is 100newton are perpendicular to each other.
If one of them makes an angle of 60newton with the resultant.
[tex]F_1=100\times \sin60=86.60\ N[/tex]
and
[tex]F_2=100\times \cos60=50\ N[/tex]
The magnitude of force,
[tex]F=\sqrt{F_1^2+F_2^2} \\\\F=\sqrt{86.6^2+50^2} \\\\F=99.99\ N[/tex]
or
F = 100 N
So, the magnitude of force is 100 N.
An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration
Answers
Answer:
218m
Explanation:
Given parameters:
Initial velocity = 3.3m/s
acceleration = 3.7m/s²
time = 10s
Unknown:
How far will it travel during the time of acceleration = ?
Solution:
We use of the kinematics equations to solve this problem;
S = ut + [tex]\frac{1}{2}[/tex] at²
S is the distance
u is the initial velocity
t is the time
a is the acceleration
So;
S = (3.3x10) + ([tex]\frac{1}{2}[/tex] x 3.7 x 10²) = 218m
A 0.5 kg apple falls from a tree. When the apple is 1.5 meters above the ground it has a velocity of 2.4 m/s. What is the apple's total mechanical energy
Answers
Answer:
8.7975 Joules
Explanation:
Step one:
Given data
mass m= 0.5kg
height h= 1.5m
velocity v= 2.4m/s
The potential energy PE= mgh
The kinetic energy KE= 1/2mv^2
Step two
PE=0.5*9.81*1.5
PE=7.3575 Joules
KE= 1/2*0.5*2.4^2
KE=0.5*0.5*5.76
KE=1.44 Joules
The total mechanical energy is
PE+KE
=7.3575+1.44
=8.7975 Joules
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cliff
Answers
Answer:
The time taken by the projectile to hit the ground is 6.85 sec.
Explanation:
Given that,
Vertical height of cliff = 230 m
Distance = 300 m
Suppose, determine the time taken by the projectile to hit the ground.
We need to calculate the time
Using second equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, s = vertical height of cliff
u = initial vertical velocity
g = acceleration due to gravity
Put the value in the equation
[tex]230=0+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]t=\sqrt{\dfrac{230\times2}{9.8}}[/tex]
[tex]t=6.85 sec[/tex]
Hence, The time taken by the projectile to hit the ground is 6.85 sec.
Needddd helppppppp!!!
Answers
Answer:
2/9 times as strong.
Explanation:
From the question given above, the following assumptions were made:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
Next, we shall obtain an expression to determine the new force. This can be obtained as follow:
F = GMm / r²
Cross multiply
Fr² = GMm
Divide both side by Mn
G = Fr² / Mm
Since G is constant, then we have
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Finally, we shall determine the new force as follow:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Fr² / Mm = F₂ × (3r)² / 2M × m
Fr² / Mm = F₂ × 9r² / 2Mm
Cross multiply
Fr² × 2Mm = F₂ × 9r² × Mm
Divide both side by 9r² × Mm
F₂ = Fr² × 2Mm / 9r² × Mm
F₂ = F × 2 / 9
F₂ = 2/9 F
Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following using the energy/work formulae only:
1.The kinetic energy at the top of the cliff
2.The total mechanical energy at the top of the cliff
3.The kinetic energy of the Rock half way down 4.the speed of the Rock half way down
5.The speed of the Rock as it hits the ground
Answers
Answer:
this is the answer......
You move a 25 N object 5 meters. If it takes 8 s how much power did you do?
Answers
Answer:
15.625 watts
Explanation:
Recall that power is defined as the worked performed per unit of time:
Power = Work / time
The work done is Force * distance, so in our case the work is:
Work = 25 M * 5 m = 125 J
Then the power will be:
Power = 125 J / 8 sec = 15.625 watts
A 1570 kg car skidding due north on a level frictionless icy road at 156 km/h collides with a 2245.1 kg car skidding due east at 120 km/h in such a way that the two cars stick together. At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at?
Answers
Answer:
θ = 47.75º East of North.
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, if we project it along E-W and N-S axes, the momentum components along these axes must be conserved too.So, for the N-S axis, we can write the following equation:[tex]p_{Northo} = p_{Northf} (1)[/tex]
Since the car moving due east has no speed component along the N-S axis, the initial momentum along this axis is simply:[tex]p_{Northo} = m_{1} * v_{1o} (2)[/tex]
where m₁ = 1570 kg, v₁₀ = 156 km/h
In order to work with the same units, we need to convert the speed in km/h to m/s, as follows:[tex]v_{1o} = 156 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 43.3 m/s (3)[/tex]
Replacing by the values in the left side of (1), we get:[tex]p_{Northo} = m_{1} * v_{1o} = 1570 kg* 43.3 m/s = 67981 kg*m/s (4)[/tex]
Since the collision is inelastic, both cars stick together, so we can write the right side of (1) as follows:[tex]p_{Northf} = (m_{1} + m_{2})* v_{fNorth} = 3815.1 kg* v_{fNorth} (5)[/tex]
From (4) and (5) , we can solve for VfNorth:[tex]V_{fNorth} = \frac{67981kg*m/s}{3815.1kg} = 17.8 m/s (6)[/tex]
We can repeat exactly the same process for the E-W axis:[tex]p_{Easto} = p_{Eastf} (7)[/tex]
Since the car moving due north has no speed component along the E-W axis, the initial momentum along this axis is simply:[tex]p_{Easto} = m_{2} * v_{2o} (8)[/tex]
As we did with v₁₀, we need to convert v₂₀ from km/h to m/s, as follows:[tex]v_{2o} = 120 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 33.3 m/s (9)[/tex]
Replacing by the values in the left side of (7), we get:[tex]p_{Easto} = m_{2} * v_{2o} = 2245.1 kg* 33.3 m/s = 74762 kg*m/s (10)[/tex]
Since the collision is inelastic, both cars stick together, so we can write the right side of (7) as follows:[tex]p_{Eastf} = (m_{1} + m_{2})* v_{fEast} = 3815.1 kg* v_{fEast} (11)[/tex]
From (10) and (11) , we can solve for VfEast:[tex]V_{fEast} = \frac{74762 kg*m/s}{3815.1kg} = 19.6 m/s (12)[/tex]
In order to find the angle East of North of the velocity vector, as we know the values of the horizontal and vertical components, we need just to apply a little bit of trigonometry, as follows:[tex]tg \theta = \frac{V_{fEast}}{V_{fNorth} } = \frac{19.6}{17.8} = 1.1 (13)[/tex]
The angle θ, East of North, is simply tg⁻¹ (1.1):θ = tg⁻¹ (1.1) = 47.75º E of N.
Electronic configuration of.
1)Fe.
2)Fe++
3)Fe+++
Answers
Answer:
The electronic configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5. Fe2+ contains 2 fewer electrons compared to the electronic configuration of Fe.
economy?
A
Supply decreases
Price decreases
B.
Pnce increases
Scaroty decreases
c.
Demand increases
Pace increases
DO
Demand increases
Price decreases
Answers
Answer:
I. don't. get. this. question
C. Demand increases
Pace increases
An endangered species is a species that has died out and no individuals are left. True or false
Answers
Answer:
the answer is false... ....
Answer:
False
Explanation:
A species that has died our ans has no individual's left would be considered extict.
A sample of 0.500 kg of ice is held at a temperature of 0°C. How much energy
must be added to completely turn the ice into liquid water? (The latent heat of
vaporization for water is 2260 kJ/kg; the latent heat of fusion for water is 333
kJ/kg.)
A. 1130 kJ
B. 666 kJ
C. 167 kJ
D. 4520 kJ
Answers
Answer:
C. 167 kJ
Explanation:
The minimum amount of heat require to complete turn the ice into liquid water is equal to the latent heat of the ice, that is, the amount of heat needed by the ice to turn into water. This amount is calculated by:
[tex]Q = m\cdot h_{f}[/tex] (1)
Where:
[tex]m[/tex] - Mass of ice, measured in kilograms.
[tex]h_{f}[/tex] - Latent heat of fusion, measured in kilojoules per kilogram.
[tex]Q[/tex] - Latent heat, measured in kilojoules.
If we know that [tex]m = 0.500\,kg[/tex] and [tex]h_{f} = 333\,\frac{kJ}{kg}[/tex], the latent heat of ice is:
[tex]Q = (0.500\,kg)\cdot \left(333\,\frac{kJ}{kg} \right)[/tex]
[tex]Q = 166.5\,kJ[/tex]
Therefore, the correct answer is C.
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a speed of 4.85 106 m/s
Answers
Answer:
8.0995×10^-21 kgms^-1Explanation:
Mass of proton :
[tex]m_P=1.67\times 10^-^2^7\:kg\\[/tex]
Speed of Proton:
[tex]v_P=4.85\times 10^6[/tex]
Linear Momentum of a particle having mass (m) and velocity (v) :
[tex]-> p =m->v\:\:\: (1)[/tex]
Magnitude of momentum :
[tex]p=mv\:\:\: (2)[/tex]
Frome equation (2), magnitude of linear momentum of the proton :
[tex]p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1[/tex]
You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 KPa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 KPa .
Find the net downward force on the tank's flat bottom, of area 2.15 m2 , exerted by the water and air inside the tank and the air outside the tank.
Answers
Answer: F = 6262.2 kN
Explanation: Pressure is defined as force per area. But pressure varies according to the depth of a fluid: in air, it decreases the higher the altitude, while in water, it increases the deeper you go.
So, at the bottom of the tank, besides the pressure of air inside the tank and air outside the tank, there is pressure of water due to its depth.
Pressure due to the depth is calculated as
[tex]P=h.\rho.g[/tex]
h is the depth in m
ρ is density of the fluid, in this case is water, so ρ = 997 kg/m³
g is acceleration due to gravity, which, in this case, is 3.71 m/s²
Then, pressure at the bottom of the tank due to variation in depth is
[tex]P=14.4(997)(3.71)[/tex]
P = 53263.73 Pa or 53.26 kPa
Assuming positive referential is downward, all pressures at the bottom point down, so total or resultant pressure is:
[tex]P_{r}=P_{1}+P_{2}+P_{3}[/tex]
[tex]P_{r}=150+88+53.26[/tex]
[tex]P_{r}=[/tex] 291.26 kPa
At last, pressure is force per area:
[tex]P=\frac{F}{A}[/tex]
[tex]F=P.A[/tex]
[tex]F_{r}=P_{r}.A[/tex]
[tex]F_{r}=291.26.10^{3}(2.15)[/tex]
[tex]F_{r}=[/tex] 626209 N or 626.2 kN
At the cylindrical tank's flat bottom, net force has magnitude 626.2 kN.
In downtown Chicago, the east-west blocks are 400 ft long while the north-south blocks are 280 ft long. Because of the many one-way streets, it can be challenging to get around. Veronica starts at the corner of Dearborn and Ohio Streets. She drives four blocks north to Superior, two blocks east to Wabash, then a block south to get to her destination at Wabash and Huron.
Required:
What is the straight-line distance from her starting point?
Answers
Answer:
The answer is "1160 ft".
Explanation:
Using the Pythagoras theorem:
[tex]\to a= 280 \times 3= 840\\\\\to b= 400 \times 2= 800\\\\[/tex]
[tex]\bold{\to y^2= a^2+b^2}\\\\[/tex]
[tex]=840^2 +800^2\\\\= 705600+640000\\\\=1345600\\\\=1160 \ ft\\[/tex]
[tex]\to y=1160 \ ft[/tex]
