Based on the calculations, the angle through which the tire rotates is equal to 4.26 radians and 244.0 degrees.
<h3>How to calculate the angle?</h3>
In Physics, the distance covered by an object in circular motion can be calculated by using this formula:
S = rθ
<u>Where:</u>
- r is the radius of a circular path.
- θ is the angle measured in radians.
Substituting the given parameters into the formula, we have;
1.87 = 0.44 × θ
θ = 1.87/0.44
θ = 4.26 radians.
Next, we would convert this value in radians to degrees:
θ = 4.26 × 180/π
θ = 4.26 × 180/3.142
θ = 244.0 degrees.
Read more on radians here: brainly.com/question/19758686
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CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
Answer:
Explanation:
v = ω R
v is linear speed and ω is angular speed
ω = v / R
a ) Inner radius = 25 x 10⁻³ m
speed v = 1.25 m/s
ω = 1.25 / (25 x 10⁻³ )
= .05 x 10⁻³
= 5 x 10⁻⁵ rad / s
outer radius = 58 x 10⁻³ m
speed v = 1.25 m/s
ω = 1.25 / (58 x 10⁻³ )
= .0215 x 10⁻³
= 2.15 x 10⁻⁵ rad / s
b )
linear constant speed v = 1.25 m /s
time = 74 min = 74 x 60 s
distance tracked = speed x time
= 1.25 x 74 x 60
= 5550 m
c ) time given
= 74 min = 74 x 60 s
angular acceleration
= ( 2.15 - 5 ) x 10⁻⁵ / (74 x 60 )
= - 6.42 x 10⁻⁹ rad / s²
I know 2
1. Temperature
2. Density of Medium