Answers
Answer:
The think the answer is solar radiation.
Explanation:
here, we gain the heat from the sun through a radiation. When it travels from the sun the harmful radiation are absorbed by ozone layer and heat enegry is provided to the surface of the Earth.
hopeit helps..
Related Questions
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.a. What is the electric field strength and direction between the plates? b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity] c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have? d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)
A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Car- tesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar coordinates?
Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?
Answers
Answer:
use google to find answer
Explanation:
69 69 69 69 69 69
Thank you!!
Answers
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kN
b. Left support = 36 kN
Right support = 29 kN
c. Left support force will decrease
Right support force will increase.
d. Right support = 43 kN
Explanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
Answers
Answer:
the googles are 5.3 m from the edge
Explanation:
Given that
depth of pool , d = 3.2 m
Now, let i be the angle of incidence
a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge
⇒tan i = 2.2/0.9
solving we get
i = 67.8°
Using snell's law ,
n1 ×sin(i) = n2 ×sin(r)
n1= refractive index of 1st medium= 1
n2= refractive index of 2nd medium = 1.33
r= angle of reflection
therefore,
r = 44.1°
Now,
distance of googles = 2.2 + d×tan(r)
distance of googles = 2.2 + 3.2×tan(44.1)
distance of googles = 5.3 m
the googles are 5.3 m from the edge
Answers
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?
Answers
Answer:
(a). Vf = 7.14 m/s
(b). Vf = 7.14 m/s
(c). same answer
Explanation:
for question (a), we would be applying conservation of energy principle.
but the initial height is h = 1.5 m
and the initial upward velocity of the ball is Vi = 10 m/s
Therefore
(a). using conservation law
Ef = Ei
where Ef = 1/2mVf² + mghf ........................(1)
also Ei = 1/2mVi² + mghi ........................(2)
equating both we have
1/2mVf² + mghf = 1/2mVi² + mghi
eliminating same terms gives,
Vf = √(Vi² + 2g (hi -hf))
Vf = √(10² + -2*9.8*2.5) = 7.14 m/s
Vf = 7.14 m/s
(b). Same process as done in previous;
Ef = Ei
but here the Ef = mghf ...........(3)
and Ei = 1/2mVi² + mghi ...........(4)
solving for the final height (hf) we relate both equation 3 and 4 to give
mghf = 1/2mVi² + mghi ..............(5)
canceling out same terms
hf = hi + Vi²/2g
hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)
recalling conservation energy,
Ef = Ei
1/2mVf² + mghf = mghi
inputting values of hf and hi we have
Vf = √(2g(hi -hf)) = 7.14 m/s
Vf = 7.14 m/s
(c). From answer in option a and c, we can see there were no changes in the answers.
a feeling of weightlessness?
Answers
Answer:
Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.
while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.
Both spacecraft and the astronauts both are in a free-fall condition.