Answer:
D
Explanation:
30 m/s is the hypothenuse. The horizontal component will be x. Using trig, you can see that
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
Answer:
The acceleration at the astronaut's head decreases.
Explanation:
Since the centripetal acceleration equals acceleration due to gravity,
a = g = GM/R². since a changes infinitesimally from his foot to his head, we differentiate a with respect to r to get da/dr = -2GM/R³.
So, da, the change in acceleration = -2GMdR/R³ = -2gdR/R = -2 × 9.8/6.4 × 10⁶ m = -3.0625 × 10⁻⁶dR m/s².
Since dR = height of astronaut = 1.80 m, da = -3.0625 × 10⁻⁶ × 1.8 = -5.5125 × 10⁻⁶ m/s².
So the acceleration at the astronaut's head is g + da = 9.8 - 0.0000055125 = 9.7999944875 m/s².
So the acceleration at the astronaut's head decreases.
Answer:
i dont think anything will happen so maybe A: the wall is larger than you.
Explanation:
Answer: equal to one
Explanation:
Because is the sum of the relative frequencies.