The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
1 hour = 3,600 seconds
1 km = 1,000 meters
75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s
The mean speed during the deceleration is
(1/2)(20-5/6 + 0) = 10-5/12 m/s.
Traveling at this average speed for 21 seconds,
the bus covers
(10-5/12) × (21) = 218.75 meters.
Response: The spring constant is 25 N/m.
Details:
The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).
The total oscillations are 20 in 4 seconds.
Oscillations per second = 
Spring's frequency of vibration is = 
The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.



The spring constant is 25 N/m.
Answer:
293.7 degrees
Explanation:
A = - 8 sin (37) i + 8 cos (37) j
A + B = -12 j
B = a i + b j, where a and b represent constants to solve for.
A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
By comparing the coefficients of i and j:
a = 8 sin (37) = 4.81452 m
b = -12 - 8cos(37) = -18.38908
Thus,
B = 4.81452 i - 18.38908 j..... 4th quadrant
<pTherefore,
cos(Q) = 4.81452 / 12
Q = 66.346 degrees
360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

Where h represents Planck’s constant (6.62*10^-34 J s).
Assuming that the electron's mass remains the same, we proceed as follows:

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:


If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:
