Answer:
A) 15.0 years
Explanation:
Due to the distance to the star system is in light-year units, we can compute the time by using:
![t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D%3D%5Cfrac%7B14.4%20l-y%7D%7B0.960%7D%3D15l-y)
then, Rob will take to complete the trip about 15 light-years.
hope this helps!!
In this item, we let x be the rate of the boat in still water and y be the rate of the current.
Upstream. When the boat is going upstream, the speed in still water is deducted by the speed of the current because the boat goes against the water. The distance covered is calculated by multiplying the number of hours and the speed.
(x - y)(3) = 144
Downstream. The speed of the boat going downstream is equal to x + y because the boat goes with the current.
(x + y)(2) = 144
The system of linear equations we can use to solve for x is,
3x - 3y = 144
2x + 2y = 144
We use either elimination or substitution.
We solve for the y of the first equation in terms of x,
y = -(144 - 3x)/3
Substitute this to the second equation,
2x + 2(-1)(144 - 3x)/3 = 144
The value of x from the equation is 60
<em>ANSWER: 60 km/h</em>
Answer:
6.14 s
Explanation:
The time the rocket takes to reach the top is only determined from its vertical motion.
The initial vertical velocity of the rocket is:
![u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s](https://tex.z-dn.net/?f=u_y%20%3D%20u%20sin%20%5Ctheta%20%3D%20%28100%29%28sin%2037%5E%7B%5Ccirc%7D%29%3D60.2%20m%2Fs)
where
u = 100 m/s is the initial speed
is the angle of launch
Now we can apply the suvat equation for an object in free-fall:
![v_y = u_y +gt](https://tex.z-dn.net/?f=v_y%20%3D%20u_y%20%2Bgt)
where
is the vertical velocity at time t
is the acceleration of gravity
The rocket reaches the top when
![v_y =0](https://tex.z-dn.net/?f=v_y%20%3D0)
So by substituting into the equation, we find the time t at which this happens:
![t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s](https://tex.z-dn.net/?f=t%3D-%5Cfrac%7Bu_y%7D%7Bg%7D%3D-%5Cfrac%7B60.2%7D%7B-9.8%7D%3D6.14%20s)
Answer:
Distance: 75 km
Displacement: 45 km
Explanation:
- Distance is a scalar quantity that refers to the total space covered by an object. It is calculated as the sum of the distances covered in each motion, regardless of their direction. therefore in this case:
distance = 60 km + 15 km = 75 km
- Displacement is a vector quantity whose magnitude is equal to the difference between the final point and the starting point of the motion, so it also takes into account the direction of each motion. In this case, the truck moves 60 km east, and then 15 km west: if we call '0' the starting point, the final point will be then
![p_B = 60 km - 15 km=45 km](https://tex.z-dn.net/?f=p_B%20%3D%2060%20km%20-%2015%20km%3D45%20km)
And so the displacement is
![d=p_B - p_A=45 km - 0 = 45 km](https://tex.z-dn.net/?f=d%3Dp_B%20-%20p_A%3D45%20km%20-%200%20%3D%2045%20km)
Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J