Answer:
11.95m/s
Explanation:
A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.
a) Find the speed of the object. Answer in units of m
K. E =½mv²
150= ½mv²
Multiply both sides by 2
mv² = 300
Divide both sides by v²
m = 300/v² .................. Equation 1
Momentum is the product of mass and velocity
Momentum = mv
25.1 = mv
Divide both sides by v
m = 25.1/v ................ Equation 2
Equate equations 1 and 2
300/v² = 25.1/v
Cross multiply
25.1v² = 300v
Multiply v with both sides
25.1v = 300
Divide both sides by 25.1
v = 300/25.1
V = 11.95m/s
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The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit
<h3>What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?</h3>
- Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ m
- Orbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)
= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
Answer:
Explanation:
We have given given the final angular velocity 
And 
Displacement 
We have to find the angular acceleration 
According to law of motion 
So 
In question we have tell about magnitude only so
