Answer:
The difference between ice and steam in Celsius (Centigrade) is 100 deg.
So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.
So 8 cm is 4 cm greater than the ice point
4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart
Then 1/5 * 100 deg C = 20 deg C the requested temperature
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
Greetings.
