Answer: C. -1.16 meters/second2
Explanation:
A= v/t (velocity/time)
in this case: v=7 and t=6
So, A= 7/6
A=1.16
The graph is decreasing so accelleration would be negative
A= <u>-1.16 meters/second2</u>
<u>Option C!</u> ; )
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From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.
<h3>What is weight?</h3>
The weight of an object is obtained as the product of the mass of the body and the acceleration due to gravity.
Thus;
When;
mass = 120 kg
weight = 46 N
acceleration due to gravity = 46 N/120 kg
=0.38 m/s^2
Learn more about acceleration due to gravity :brainly.com/question/13860566
#SPJ1
Answer:
2.62seconds
Explanation:
Speed is defined as the ratio of the distance covered by a body with respect to time.
Speed v = Distance (s)/Time (t)
For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.
Speed v = 2 × distance (S)/Time (T)
V = 2S/t
2S = vt
Given speed of the wave = 342m/s
Distance covered = 450m
t = 2S/v
t = (2×450)/343
t = 900/343
t = 2.62seconds
It will take him 2.62seconds for him to hear his own voice echo off of the wall.
Answer:
The water level rises more when the cube is located above the raft before submerging.
Explanation:
These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.
Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.
When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.
Vc = 0.45*0.45*0.45 = 0.0911 [m^3]
Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.
![Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]](https://tex.z-dn.net/?f=Ro_%7BH2O%7D%3DR_%7Bc%2Br%7D%5C%5Cwhere%3A%5C%5CRo_%7BH2O%7D%3D%20water%20density%20%3D%201000%20%5Bkg%2Fm%5E3%5D%5C%5CRo_%7Bc%2Br%7D%3D%20combined%20density%20cube%20%2B%20raft%20%5Bkg%2Fm%5E3%5D)
Density is given by:
Ro = m/V
where:
m= mass [kg]
V = volume [m^3]
The buoyancy force can be calculated using the following equation:
![F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]](https://tex.z-dn.net/?f=F_%7BB%7D%3DW%3DRo_%7BH20%7D%2Ag%2AVs%5C%5CW%20%3D%20%28200%2B730%29%2A9.81%5C%5CW%3D9123.3%5BN%5D%5C%5C%5C%5C9123%3D1000%2A9.81%2AVs%5C%5CVs%20%3D%200.93%20%5Bm%5E3%5D)
Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.
The new volume = 3 x 52.6 that’s because as the pressure decreases by 1/3 the volume increases x3