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Bad White [126]
2 years ago
11

Under which one of the following circumstances will heat transfer occur via convection? Group of answer choices Convection occur

s within metal objects. Convection only occurs in non-metallic solids. Convection occurs only within a vacuum. Convection occurs in the presence of a liquid or a gas. Convection can occur whether matter is present or not.
Physics
2 answers:
BartSMP [9]2 years ago
8 0

Answer:

1) in metal object heat transfer through the conduction .In vacuum heat transfer through only radiation .

In only gaseous state or in liquid state the heat transfer through convection hence option D is correct

Sever21 [200]2 years ago
8 0

Answer:

The answer is: Convection occurs in the presence of a liquid or a gas

Explanation:

Convection is a process by which heat is transmitted, either between two liquid or gaseous substances, provided they are at different temperatures.

Heat transfer occurs when there is a temperature gradient between the two bodies that are in direct contact; then, the heat flow flows from the hottest body to the least hot body. Liquids and gases are considered fluids.

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A nervous squirrel gets startled and runs 5.0\,\text m5.0m5, point, 0, space, m leftward to a nearby tree. The squirrel runs for
mel-nik [20]

Answer:

−5.0  

Explanation:

6 0
1 year ago
Read 2 more answers
A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds
horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

4 0
2 years ago
Read 2 more answers
How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by th
ololo11 [35]

Answer:

1.0\cdot 10^{41} times

Explanation:

First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.

Age of the universe:

T=100,000,000,000,000,000s = 1.0\cdot 10^{17} s (1 followed by 17 zeroes)

Lifetime of the top quark:

\tau = 0.000000000000000000000001s = 1.0\cdot 10^{-24} s (we moved the decimal point 24 places to the right)

Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

r = \frac{T}{\tau}=\frac{1.0\cdot 10^{17} s}{1.0\cdot 10^{-24} s}=1.0\cdot 10^{41}

6 0
2 years ago
Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi
aleksklad [387]
1) The buoyant force acting on an object immersed in a fluid is:
B=d_f V_d g
where d_f is the density of the fluid, V_d is the volume of displaced fluid, and g=9.81~m/s^2 is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3. Using the relationship between density, volume and mass, we find
V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is d_f = 1~Kg/m^3, so we have
B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N

4) The weight of 1 Kg of titanium is instead:
W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N
So, the buoyant force is negligible compared to the weight.
7 0
1 year ago
Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m
Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

A=\frac{1-V_{1}^{2}  }{E_{1} }-\frac{1-V_{2}^{2}  }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}

a) The maximum contact pressure is:

P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} })  }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa

The width of contact is:

b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} })  } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0
2 years ago
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