The weight could be different, metals have a higher mass than nonmetals, so when occupying the same amount of space, the weight of the metal is far more.
I think is A or B it depends on like what the trying to answer
Answer:
17 °C
Explanation:
From specific Heat capacity.
Q = cm(t₂-t₁)................. Equation 1
Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.
make t₁ the subject of the equation
t₁ = t₂-(Q/cm)............... Equation 2
Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c
Substitute into equation 2
t₁ = 22-[5000/(4×250)
t₁ = 22-(5000/1000)
t₁ = 22-5
t₁ = 17 °C
You need to check the temperature of food being stored in a temperature-controlled environment every four hours. The process of changing a space's temperature is called temperature control.
Cooking food alone may not be enough to avoid food poisoning, though, if the bacteria in food are allowed to grow to large numbers. When the temperature is between 5°C and 63°C, bacteria can grow. The risk zone is the range between 5°C and 63°C.
Temperature control is a process where the passage of heat energy into or out of a space or substance is adjusted to achieve the desired temperature. This process involves measuring or otherwise detecting changes in the temperature of the space (and all of the objects contained therein) or of the substance.
Learn more about temperature here
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Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
