Since I've been waiting to get my answers checked for my submission #2, I moved on to my submission #3 and I need some questions checked for my submission #3 please.
1.)If a blue car drives 10km/h across the deck of an ocean liner that is traveling at 90km/h, what is the blue car's speed relative to the surface of the water when it drives toward the front of the ocean liner?
a)10 km/h
b)80 km/h
c)90 km/h
=d)100 km/h
2.)A motorboat travels at a speed of 40 km/h relative to the surface of the water. If it travels upstream against a current of 12 km/h what is the boats speed relative to the shore?
a)3.3 km/h
=b)28 km/h
c)40 km/h
d)52 km/h
3.)An airplane flies north at a speed of 150 km/h relative to the air through which it flies. If there is a tail wind of 50 km/h, what is the plane's velocity relative to the ground?
a)3 km/h north
=b)100 km/h north
c)150 km/h north
d)200 km/h north
4.)A plane flying 100km/h due north encounters a crosswind of 100km/h "from the east" (i.e. blowing westward). What then, is the plane's velocity relative to the ground?
a)141.4 km/h northeast
=b)141.4 km/h northwest
c)100 km/h northeast
d)100 km/h northwest
5.)A bullet is shot from a rifle aimed horizontally at the same time a bullet is dropped from the same height as the gun's barrel. Which bullet hits the ground first?
a)The fired bullet
=b)The dropped bullet
c)Both bullets the ground at the same time
6.)A car drives off a vertical cliff that is 122.5m tall. If the car lands 100m from the base of the cliff what was the cars speed as it drove off the cliff.
a)20m/s
=b)22m/s
c)24m/s
d)26m/s
(#6 I just really guessed because I don't know what to do so I just 122.5-100=22.5 so I just stuck with that.)
#1 correct
#2 correct
#3 wrong
#4 correct
#5 wrong
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#6
find t in seconds:
122.5m = (0.5)(g)(t^2)
122.5m = (0.5)(9.8m/s^2)t^2
reducing...
25s^2 = t^2
t = +5 seconds and t = -5 seconds
Disregard the negative time.
That is how long it took to fall 122.5m.
So, if the car traveled horizontally 100m in 5 seconds, speed is 100m/5s = 20m/s which is the answer a.
A more complete explanation...
The car is traveling horizontally at a constant speed. When it drives off the cliff, it keeps going at the same speed (disregarding friction) until it is stopped by hitting the ground. A forumla for (d) distance traveled with constant acceleration, no initial vertical velocity is:
d = (1/2)(a)(t^2)
where a is the acceleration and t is the time. The acceleration due to gravity is about 9.8m/s^2.
d is given as 122.5m.
Once the falling time is found that can be plugged into the formula for speed.
In this case it took 5 seconds to fall, and the car traveled 100m in that time.
In number 5 both bullets fall at the same rate. The horizontal problem has nothing to do with the vertical problem except that the time in the air is the same.
Thanks:)
A motorboat travels at a speed of 40 km/h relative to the surface of the water. If it travels upstream against a current of 12 km/h what is the boats speed relative to the shore?
An airplane flies north at a speed of 150 km/h relative to the air through which it flies. If there is a tail wind of 50 km/h, what is the plane's velocity relative to the ground?
.)An airplane flies north at a speed of 150 km/h relative to the air through which it flies. If there is a tail wind of 50 km/h, what is the plane's velocity relative to the ground?
200 km/h north
.)An airplane flies north at a speed of 150 km/h relative to the air through which it flies. If there is a tail wind of 50 km/h, what is the plane's velocity relative to the ground?
150 km/h
who really cares?
To solve these questions, we need to understand the concept of relative velocities.
1.) To find the blue car's speed relative to the surface of the water when it drives towards the front of the ocean liner, we need to consider the sum of the car's speed and the liner's speed. In this case, the car's speed relative to the water would be 10 km/h (its own speed) + 90 km/h (liner's speed) = 100 km/h. Therefore, the correct answer is option d) 100 km/h.
2.) To find the boat's speed relative to the shore when it travels upstream against a current, we need to consider the difference between the boat's speed and the current's speed. In this case, the boat's speed relative to the shore would be 40 km/h (its own speed) - 12 km/h (current's speed) = 28 km/h. Therefore, the correct answer is option b) 28 km/h.
3.) To find the plane's velocity relative to the ground when there is a tailwind, we need to consider the sum of the plane's speed and the tailwind's speed. In this case, the plane's velocity relative to the ground would be 150 km/h (its own speed) + 50 km/h (tailwind's speed) = 200 km/h. Therefore, the correct answer is option d) 200 km/h north.
4.) To find the plane's velocity relative to the ground when there is a crosswind, we can use the Pythagorean theorem. The plane's velocity vector relative to the ground can be represented as the sum of its velocity vector due north and its velocity vector due west (opposite direction of the crosswind). Applying the Pythagorean theorem, the magnitude of the resulting vector is approx. 141.4 km/h. The direction of the resulting vector would be northwest. Therefore, the correct answer is option b) 141.4 km/h northwest.
5.) Both bullets will hit the ground at the same time. The horizontal motion (forward) of the bullet shot from the rifle does not affect the vertical motion (downward motion due to gravity). Both bullets experience the same acceleration due to gravity, and the time taken to hit the ground is only determined by the vertical motion. Therefore, the correct answer is option b) The dropped bullet.
6.) To find the car's speed as it drives off the cliff, we can use the equations of motion. The car's initial vertical velocity can be assumed to be zero since it starts from rest. The distance fallen is 122.5 m. We can use the equation of motion s = ut + 0.5at^2, where s is the distance fallen, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearranging the equation, we get the time t = √(2s/a). Plugging in the values, we get t = √(2 * 122.5 / 9.8) ≈ 5 seconds. Now, we can use the equation of motion v = u + at to find the final velocity. Plugging in the values, we get v = 0 + 9.8 * 5 = 49 m/s (downward). Therefore, the correct answer is option b) 22 m/s.