Working displayed in the picture below, the answer is -11 m s^-2
1) What describes the unusually large release of plasma from the sun's corona?
1 answer:
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The frequency of a simple harmonic oscillator such as a spring-mass system is given by
where
k is the spring constant
m is the mass attached to the spring.
Re-arranging the formula, we get:
and since we know the constant of the spring:
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
where
k is the spring constant
m is the mass attached to the spring.
Re-arranging the formula, we get:
and since we know the constant of the spring:
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
