Answer:
C
Explanation:
A food chain outlines who eats whom. A food web is all of the food chains in an ecosystem.
Answer:
C
Hope this helped :D
A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Answer:
T = 98 N
Explanation:
The gravity of the earth is known to be 9.8 m/s²
Data:
m = 10 kgg = 9.8 m/s²T = ?Use formula:
[tex]\boxed{\bold{T=m*g}}[/tex]Replace and solve:
[tex]\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{T=98\ N}}}[/tex]The tension in the rope is 98 Newtons.
Greetings.
A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid inside a control surface and determine the change in internal energy, in kJ, of this control mass.
Answer: [tex]3590\ kJ[/tex]
Explanation:
Given
Paddle wheel work is [tex]W=-5090\ kJ\quad \text{work is done on the system}[/tex]
Heat transfer from the tank is [tex]Q=-1500\ kJ\quad \text{heat taken from the system}[/tex]
From the first law of thermodynamics
Change in the internal energy of the system is equal to the difference of heat and work .
[tex]\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ[/tex]
Therefore, the change in internal energy is [tex]3590\ kJ[/tex]
A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.
Answer:
272° C
Explanation:
Given :
Volume of the balloon, V = 500 [tex]m^3[/tex]
The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]
Total load, [tex]m_{T}[/tex] = 290 kg
Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]
We known buoyant force,
[tex]$F_B = \rho_{air} V$[/tex]
For a 290 kg lift, [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]
[tex]$m=\rho V$[/tex]
∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]
[tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]
[tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]
[tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]
∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]
[tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]
[tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]
= 545 K
[tex]$=272^\circ C$[/tex]
Therefore, temperature of the air in the balloon is 272 degree Celsius.
To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.
The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.Reasons:
Given information are;
Volume of the balloon = 500.0 m³
Temperature of the surrounding air = 15.0°C
Density of air at 15.0°C = 1.23 kg/m³
Required:
The temperature required to lift 290kg.
Solution:
Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;
[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500
Therefore;
[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]
According to the Ideal Gas Law, we have;
ρ₁ × R × T₁ = ρ₂ × R × T₂
Therefore;
[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]
Therefore;
[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]
The temperature of the balloon, T₂ ≈ 545.27 - 273.15 = 272.12
The temperature of the air in the balloon, T₂ ≈ 272.12 °C
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kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay
Explanation:
4. Alin sa mga sumusunod na awitin ang may tempong presto?
a. “Chua-ay”
c. “Akong Manok”
b. “Sitsiritsit
d. "Ili-ili Tulog Anay”
5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?
a. mabagal
c. mabilis na mabilis
b. mabilis at mabagal d. katamtamang bilis
6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?
a. descant
b. ostinato
c. tempo
d. texture
7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?
a. alto
b. forte
c. tempo
d. soprano
8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?
a. alto
b. forte
c. tempo
d. soprano
9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?
a. descant
b. ostinato
c. melodic ostinato
d. rhythmic ostinato
10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?
a. descant
b. ostinato
c. melodic ostinato
d. rhythmic ostinato
A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, what type of path will the particle follow
Answer:
a circular path
Explanation:
In a magnetism field if a charged particle having a charge of magnitude '' enters such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F
[tex]F = V\times B[/tex]
According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
[tex]Weight\ loss=1.6321N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=85.9kg[/tex]
Altitude [tex]h= 6.33 km[/tex]
Let
Radius of Earth [tex]r=6380km[/tex]
Gravity [tex]g=9.8m/s^2[/tex]
Generally the equation for Gravity at altitude is mathematically given by
[tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]
[tex]g_s=9.781m/s^2[/tex]
Therefore
Weight at sea level
[tex]W_s=9.8*85.9[/tex]
[tex]W_s=841.82N[/tex]
Weight at 6.33 altitude
[tex]W_a=9.781*85.9[/tex]
[tex]W_a=840.2N[/tex]
Therefore
[tex]Weight loss=W_s-W_b[/tex]
[tex]Weight loss=841.82-840.2[/tex]
[tex]Weight loss=1.6321N[/tex]
a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.
PLEASE HELP I NEED THIS TODAY
Answer:
a)I=V/R
39.5 amp
Explanation:
because the voltage in serious with 1ohm resistor
You’re working with a patient who suddenly falls. You should?
is anyone online??just asking
Answer:
me...:(
Explanation:
Answer:
hello I'm online here thanks for the points (◔‿◔)
Which of these is a source of thermal energy inside earth
There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical speed be right before it hits the ground?
A. 0 m/s
B. 15 m/s
C. 40 m/s
D. 30 m/s
Answer:
Explanation:
The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
Answer:
50 watts
Explanation:
Applying,
Power (P) = Workdone (W)/Time(t)
But,
Work done (W) = Force (F)×distance(d)
Therefore,
P = Fd/t..................... Equation 1
Where P = power of the weightlifter, F = Force applied, d = distance, t = time.
From the question,
Given: F = 200 N, d = 0.5 m, t = 2 s
Substitute these values into equation 1
P = (200×0.5)/2
P = 100/2
P = 50 watts
At what speed was object A moving ?
Answer:
C
Explanation:
The answer is C because if you look at the 1 hour mark it shows 10km
Answer:It will be 10km/hour
Explanation:
The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N
Answer:
8760 N
Explanation:
think this is the right answer :)
If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright
A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?
a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved
The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade
Answer:
The centripetal acceleration is 6.95 m/s²
Explanation:
Given;
angular displacement of the blade, θ = 90.08⁰
duration of motion of the blade, t = 0.4 s
radius of the circle moved by the blade, r = 0.45 m
The angular speed of the blade in radian is calculated as;
[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]
The centripetal acceleration is calculated as;
a = ω²r
a = (3.93)² x 0.45
a = 6.95 m/s²
What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine
Which of the following would likely happen if a person’s lactic acid system had difficulty breaking down glycogen in the muscles?
The person would have difficulty swimming across a lake.
The person would have difficulty sprinting in a race.
The person would have difficulty cycling down a hill.
The person would have difficulty running a marathon.
Answer: I think that its b, they would have difficulty sprinting in a race
Explanation:
If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process. Express your answer in volts.
The question is incomplete. The complete question is :
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.
Solution :
Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.
It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.
Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :
[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex] and [tex]$\phi_{B,f} = 0$[/tex]
Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,
[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]
[tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]
= 30.27 V
Therefore, the emf generated is 30.27 V.
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I’ll make you the Brainlyest, I can’t get this one wrong.
A car drives 110 km in 2 hours. Calculate the speed of the car
Answer: 55 kmph
Explanation: Divide 110 by 2
А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity
Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 48 s. What is Jason's top speed
Answer:
81.1 m/s
Explanation:
The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration
So, T - f = ma
T - μmg = ma
a = T/m - μg
susbstμituting the values of the varμiables into the equation, we have
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 2.67 m/s² - 0.98 m/s²
a = 1.69 m/s²
Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s
So, v = u + at
v = 0 m/s + 1.69 m/s² × 48 s
v = 0 m/s + 81.12 m/s
v = 81.12 m/s
v ≅ 81.1 m/s
So, Jason's top speed is 81.1 m/s
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
please help! will mark brainliest
Answer:
27.D.28.AExplanation:
THE ANSWERME MY ANSWERWAH IS ME AND MY ANSWER WAHHHWALA NA SIYA WALA NA SIYA MASAYA MASAYA MASAYA WAHHtinapon na siya tipon na siya wahhsa basurahan wahh wahh ...............If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
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Convert 387.1 K to °C
A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?
Answer:
Answer is 183.6 J
Explanation:
Using the Physics reference sheet the formula for Potential energy is
(mass) x (gravity) x (height)
Mass= 1.2
Gravity I used is 9.81 (use 10 to get the answer most schools use)
Height= 15.6
What are impact and non-impact printers?
Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.
In Non-Impact printers, no mechanical moving component is used.
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