The heat capacity of the silver spoon at the given temperature difference is 817.65 J.
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Heat capacity of the silver spoon</h3>
The heat capacity of theb silver spoon is the quantity of heat absorbed by the silver spoon. The heat capacity of the silver spoon at the given temperature difference is calculated as follows;
Q = mcΔθ
where;
- m is mass of the spoon
- c is specific heat capacity of silver = 0.237 J/g⁰C
- Δθ is change in temperature
Q = 50 x 0.237 x (89 - 20)
Q = 817.65 J
Thus, the heat capacity of the silver spoon at the given temperature difference is 817.65 J.
Learn more about heat capacity here: brainly.com/question/16559442
Answer:
U = √Rg/sin2θ
Explanation:
Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.
Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.
According to the range formula,
R = U²sin2θ/g
Cross multiplying we have;
Rg = U²sin2θ
Dividing both sides by sin2θ, we have;
U² = Rg/sin2θ
Taking the square root of both sides we have;
√U² = √Rg/sin2θ
U = √Rg/sin2θ
Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ
Answer:
The magnitude of the acceleration of the tip of the minute hand of the clock 
.
Explanation:
Given that,
Length of minute hand = 0.55 m
Length of hour hand = 0.26 m
The time taken by the minute hand to complete one revelation is
We need to calculate the angular frequency
Using formula of angular frequency
Put the value into the formula
We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock
Using formula of acceleration
Put the value into the formula
Hence, The magnitude of the acceleration of the tip of the minute hand of the clock .
A material that has high resistance to the flow of electric current is called an electric resistor
