Answer:
2361 Newtons
Explanation:
From the second Newton's law of motion;
F = ma
In this case;
we are given;
Mass as 9.5 g
Initial speed as 0 m/s
Final velocity as 650 m/s
Distance is 0.85 m
Using the equation;
V² = U² + 2as
But u = 0
v² = 2as
Therefore;
a = v² ÷ 2s
= 650² ÷ 2(0.85)
= 248,529.40 m/s²
But;
F = ma
= 0.0095 kg × 248,529.40 m/s²
= 2361 Newtons
Therefore;
The average net force required to accelerate the bullet is 2361 Newtons.
Explanation:
|F|= B/2
B/2 = √(A²+B²+2ABcosθ) ------(1)
Since the resultant of A and B is perpendicular to Vector A
tan90°= BSinθ/(A+BCosθ)
(A+BCosθ)=0
Cosθ=-A/B ----(2)
Using equation (1)
B/2 = √(A²+B²+2ABcosθ)
B/2 = √(A²+B²+2AB×-A/B)
B/2=√(A²+B²-2A²)
B/2=√(B²-A²)
B²/4=B²-A²
A²=B²-B²/4
A²=3B²/4
A=√3B/2
Using equation (2)
Cosθ=-A/B
Cosθ=-[√3B/2]/B
Cosθ=-√3/2
θ= cos^-1 (-√3/2)
θ= 150°
Sound waves travel faster
Answer:
<u>
</u> is the centripetal acceleration.
Explanation:
As per given values
Radius of earth (r) = 6371000 m
The "international space station" is orbiting with a velocity (v) = 7667 m/s.
"Centripetal acceleration" is the acceleration is equal to "the square of the velocity" divided by "the radius of the circular path".
![\text { Centripetal acceleration a }=\frac{V^{2}}{R}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Centripetal%20acceleration%20a%20%7D%3D%5Cfrac%7BV%5E%7B2%7D%7D%7BR%7D)
V = velocity of the orbit
R = radius of the earth + height of the space station
R = 6,371,000 + 408,000
R = 6779000 m
The direction of the centripetal acceleration is always inwards along the radius vector of the circular motion.
![a=\frac{7667^{2}}{6779000}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B7667%5E%7B2%7D%7D%7B6779000%7D)
![a=\frac{58782889}{6779000}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B58782889%7D%7B6779000%7D)
![\mathrm{a}=8.671 \mathrm{m} / \mathrm{s}^{2}](https://tex.z-dn.net/?f=%5Cmathrm%7Ba%7D%3D8.671%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D)
![\text { The International Space Station's centripetal acceleration } 8.671 \mathrm{m} / \mathrm{s}^{2}.](https://tex.z-dn.net/?f=%5Ctext%20%7B%20The%20International%20Space%20Station%27s%20centripetal%20acceleration%20%7D%208.671%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D.)
Answer:
7.1 J
Explanation:
From the question,
Work done by the mover = work done in pushing the crate + work done against friction
W = W'+Wf................. Equation 1
W = mgd+mgμd............ Equation 2
W = mgd(1+μ)................ Equation 3
Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.
Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5
constant: g = 9.8 m/s²
Substitute these values into equation 3
W = 46×9.8×0.0105(1+0.5)
W = 7.1 J