Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
145.8m
Explanation:
The toss distance is given by:
Current will be
now just pluf in the values and Voila..
Answer:
Explanation:
The question is incomplete.
The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.
s= sin2(pi)t
Acceleration = d²S/dt²
dS/dt = 2πcos2πt
d²S/dt² = -4π²sin2πt
A(t) = -4π²sin2πt
Next is to find acceleration after 4.5 seconds
A(4.5) = -4π²sin2π(4.5)
A(4.5) = -4π²sin9π
A(4.5) = -4π²sin1620
A(4.5) = -4π²(0)
A(4.5) = 0m/s²
Answer:
See below ~
Explanation:
An object will sink in water when its density is greater than that of water, which is 1 g/cm³.
Volume of the box is <u>1331 cm³</u>. (11³)
Maximum mass of sand will be 1331 g. [because 1331/1331 = 1 g/cm³]
- Volume of sand = Mass of sand / Density of sand
- Volume (sand) = 1331/3.5
- Volume (sand) = 380.29 cm³
If the volume of sand is <u>greater than 380.29 cm³</u>, the box will sink in water.
