If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.
A, and D are two correct answers im not sure about the third sorry
Answer:
13.5 ms
Explanation:
The sound wave travels with uniform motion in both cases, so the time taken to cover a distance d is given by:
![t=\frac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D)
where
d is the distance to cover
v = 340 m/s is the speed of sound in air
So for the people across the newsroom,
d = 4.7 m
So the time taken is
![t=\frac{4.7}{340}=0.0138 s = 13.8 ms](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B4.7%7D%7B340%7D%3D0.0138%20s%20%3D%2013.8%20ms)
The radio waves instead travels with uniform motion at the speed of light:
![c=3.0\cdot 10^8 m/s](https://tex.z-dn.net/?f=c%3D3.0%5Ccdot%2010%5E8%20m%2Fs)
So the time taken for them is
![t=\frac{d}{c}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bc%7D)
where
d = 82 km = 82,000 m
is the distance of the people who are 82 km away. Substituting,
![t=\frac{82,000}{3.0\cdot 10^8}=0.27\cdot 10^{-3} s = 0.27 ms](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B82%2C000%7D%7B3.0%5Ccdot%2010%5E8%7D%3D0.27%5Ccdot%2010%5E%7B-3%7D%20s%20%3D%200.27%20ms)
Therefore, the difference in time is
![\Delta t = 13.8 ms - 0.27 ms = 13.5 ms](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%2013.8%20ms%20-%200.27%20ms%20%3D%2013.5%20ms)
Explanation:
The object's weight on Earth is 5.0 N, so its mass is:
F = ma
5.0 N = m (10 m/s²)
m = 0.50 kg
The acceleration on the moon is g/6 = (10 m/s²) / 6 = 1.67 m/s².
The velocity it reaches after 3.0 seconds is:
v = at + v₀
v = (1.67 m/s²) (3.0 s) + (0 m/s)
v = 5 m/s
So the momentum is:
p = mv
p = (0.50 kg) (5 m/s)
p = 2.50 kg m/s
Answer:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Explanation:
From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration (
), measured in meters per square second, is entirely centripetal and is expressed as:
(1)
Where:
- Period of rotation, measured in seconds.
- Radius of rotation, measured in meters.
If we know that
and
, then the time taken by the object to complete one revolution is:
![T^{2} = \frac{4\pi^{2}\cdot R}{a}](https://tex.z-dn.net/?f=T%5E%7B2%7D%20%3D%20%5Cfrac%7B4%5Cpi%5E%7B2%7D%5Ccdot%20R%7D%7Ba%7D)
![T = 2\pi\cdot \sqrt{\frac{R}{a} }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%5Ccdot%20%5Csqrt%7B%5Cfrac%7BR%7D%7Ba%7D%20%7D)
![T = 2\pi\cdot \sqrt{\frac{0.93\,m}{26.36\,\frac{m}{s^{2}} } }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%5Ccdot%20%5Csqrt%7B%5Cfrac%7B0.93%5C%2Cm%7D%7B26.36%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%7D%20%7D)
![T \approx 1.180\,s](https://tex.z-dn.net/?f=T%20%5Capprox%201.180%5C%2Cs)
The object takes approximately 1.180 seconds to complete one horizontal circle.