One of two things is true about this question: EITHER it can't happen
as you've described it, OR you've left out some vital information.
-- IF the first stone was thrown downward with an initial speed and the
second one was dropped from rest 1 second later, then the second one
can never catch up with the first one, and they can never hit the water together.
-- IF the first stone was thrown downward with an initial speed, AND the
second one was released 1 second later, AND they actually do hit the
water together, THEN the second stone must have been given an initial
downward speed greater than 2 m/s, otherwise it could never catch up
with the first one.
Note:
The masses and weights of the stones are irrelevant and not needed.
=======================================================
An afterthought . . . . .
If the first stone was tossed UP at 2 m/s . . . that could be the meaning of the
prominent plus-sign that you wrote next to the 2 . . . then it rises for (2/9.8) second, then begins to fall, and passes the mountain climber's hand on the way down (4/9.8) second after he tossed it, falling at the same 2.0 m/s downward.
From there, it still has 50m to go before it hits the water.
50 = 2 T + 1/2 G T²
4.9 T² + 2 T - 50 = 0
T = 3 seconds
The first stone hits the water 3 seconds after passing the mountain climber's hand on the way down at a downward speed of 2.0 m/s. In that 3 seconds, it gains (3 x 9.8) = 29.4 m/s of additional speed, hitting the water at (29.4 + 2) = 31.4 m/s .
This is all just a guess, assuming that the 2.0 m/s was an UPWARD launch.
Maybe I'll come back later and calculate the second stone.
Answer:
Radio waves are electromagnetic waves
Explanation:
•Radio waves are electromagnetic waves.
•Satellites are located within our atmosphere.
•Radio waves are very high energy mechanical waves.
•Radio waves are changed to gamma rays for transmission.
Answer:
The car overtakes the truck at a distance d = 3266.2ft from the starting point
Explanation:
Problem Analysis
When car catches truck:
dc = dt = d
dc: car displacement
dt: truck displacement
tc = tt = t
tc: car time
tt : truck time
car kinematics :
car moves with uniformly accelerated movement:
d = vi*t + (1/2)a*t²
vi = 0 : initial speed
d = (1/2)*a*t² Equation (1)
Truck kinematics:
Truck moves with constant speed:
d = v*t Equation (2)
Data
We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .
Development problem
Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)
Equation (1) = Equation (2)
(1/2)*a*t² = v*t
(1/2)*3*t² = 70*t (We divide both sides by t)
1.5*t = 70
t = 70 ÷ 1.5
t = 46.66 s
We replace t = 46.66 s in equation (2) to calculate d:
d = 70*46.66 = 3266.2ft
d = 3266.2 ft
Answer:
3,97078708130496x10^16 meters.
Explanation:
First of all you have to know the light speed which is approximately .
Next you have to know the distance formula that is d = s * t. d stands for distance, s stands for speed, and t stands for time.
Now you have the speed of light that was given above, but, you have time in years and need it to be in seconds to be suppressed with the speed seconds. So we calculate the time in seconds.
For 1 day, you have 24 hours, that's equal to 1440 minutes, that's equal to 86400 seconds. So, using a simple three rule if 1 day is equal to 86400 seconds, 365 days is equal to 31536000 seconds.
Now, you multiply that result ( 31536000 seconds) times 4.2 (Which is the time that takes it to reach the Earth), and you get 132451200 seconds.
Next step is to simply replace the values in the formula, and do the multiplication.
Have a nice day.
80+20 is 100
you would not count the part where the dog runs back to its master
so its 100 meters