Answer:
Here,
Initial velocity(u)=0 m/s
acceleration(a)=1.5m/s
time(t)=0.5s
Now,
distance covered(s)=ut+(1/2)at^2
=0*0.5+(1/2)*1.5*0.5*0.5
=0+(1/2)*0.375
=0+0.1875
=0.1875(nearly 0.19)
Hence,0.19 is correct answer.
Answer:
a) Time = 2.67 s
b) Height = 35.0 m
Explanation:
a) The time of flight can be found using the following equation:
(1)
Where:
: is the final position in the horizontal direction = 80 m
: is the initial position in the horizontal direction = 0
: is the initial velocity in the horizontal direction = 30 m/s
a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)
t: is the time =?
By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:
b) The height of the hill is given by:
Where:
: is the final position in the vertical direction = 0
: is the initial position in the vertical direction =?
: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)
g: is the acceleration due to gravity = 9.81 m/s²
Hence, the height of the hill is:
I hope it helps you!
Average acceleration is the ratio of the change in velocity to the time it takes for that change to occur:
We want to find 
:
<span>17%
Any object that is floating will display the volume of fluid that matches the mass of the object that's floating. So if the object is 100% as dense as the fluid, it will just barely sink, if it's 50% as dense as the fluid, 50% will be submerged, etc. So let's see what percent of mercury's density is concrete.
2.3x10^3 / 13.6x10^3 = 0.169117647 = 16.9117647%
Rounding to 2 significant figures gives 17%.
So 17% of a piece of concrete will be submerged in mercury when it's floating in mercury.</span>
Answer:
The answer to your question is given below
Explanation:
Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.
From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.
Remember, both objects must get to the ground at the same time..!
Let the time taken for both objects to get to the ground be t.
Time A = Time B = t
But B falls through time 2t
Therefore,
Time A = Time B = 2t
Height = 1/2gt^2
For A:
Time = 2t
dA = 1/2 x g x (2t)^2
dA = 1/2g x 4t^2
For B
Time = t
dB = 1/2 x g x t^2
Equating dA and dB
dA = dB
1/2g x 4t^2 = 1/2 x g x t^2
Cancel out 1/2, g and t^2
4 = 1
4dA = dB
Divide both side by 4
dA = 1/4 dB
