Answer:
Density of unknown material = 3.86 g/cm³
Explanation:
Mass of unknown material, M = 54 gm
The object occupied 14 gm of water.
Volume of unknown material = Volume of 14 gm of water.
Density of water = 1 gm/cm³
Volume of 14 gm of water
![V=\frac{14}{1}=14cm^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B14%7D%7B1%7D%3D14cm%5E3)
Volume of unknown material, V = Volume of 14 gm of water = 14 cm³
Density of unknown material
Complete Question:
A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level
a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.
b) What is the height, hn in meters?
Answer:
a) Energy = mghₙ
b) Height, hₙ = 5.02 m
Explanation:
a) Total energy in terms of maximum height
Let maximum height be hₙ
At maximum height, velocity, V=0
Total mechanical energy , E = mgh + 1/2 mV^2
Since V=0 at maximum height, the total energy in terms of maximum height becomes
Energy = mghₙ
b) Height, hₙ in meters
mghₙ = mgh + 1/2 mV^2
mghₙ = m(gh + 1/2 V^2)
Divide both sides by mg
hₙ = h + 0.5 (V^2)/g
h = 2.15m
g = 9.8 m/s^2
V = 7.5 m/s
hₙ = 2.15 + 0.5(7.5^2)/9.8
hₙ = 2.15 + 2.87
hₙ = 5.02 m
Explanation:
An electron is released from rest, u = 0
We know that charge per unit area is called the surface charge density i.e. ![\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7Bq%7D%7BA%7D%3D2.1%5Ctimes%2010%5E%7B-7%7D%5C%20C%2Fm%5E2)
Distance between the plates, ![d=1.2\times 10^{-2}\ m](https://tex.z-dn.net/?f=d%3D1.2%5Ctimes%2010%5E%7B-2%7D%5C%20m)
Let E is the electric field,
![E=\dfrac{\sigma}{\epsilon_o}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B%5Csigma%7D%7B%5Cepsilon_o%7D)
![E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B2.1%5Ctimes%2010%5E%7B-7%7D%7D%7B8.85%5Ctimes%2010%5E%7B-12%7D%7D)
E = 23728.81 N/C
Now, ![ma=qE](https://tex.z-dn.net/?f=ma%3DqE)
![a=\dfrac{qE}{m}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BqE%7D%7Bm%7D)
![a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%2023728.81%7D%7B9.1%5Ctimes%2010%5E%7B-31%7D%7D)
![a=4.17\times 10^{15}\ m/s^2](https://tex.z-dn.net/?f=a%3D4.17%5Ctimes%2010%5E%7B15%7D%5C%20m%2Fs%5E2)
Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :
![v^2=2ad](https://tex.z-dn.net/?f=v%5E2%3D2ad)
![v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}](https://tex.z-dn.net/?f=v%5E2%3D2%5Ctimes%204.17%5Ctimes%2010%5E%7B15%7D%5Ctimes%201.2%5Ctimes%2010%5E%7B-2%7D)
![v=10.003\times 10^6\ m/s](https://tex.z-dn.net/?f=v%3D10.003%5Ctimes%2010%5E6%5C%20m%2Fs)
Hence, this is the required solution.
A push or a pull on an object
Answer:
1.) Displacement = 32.08m
2.) Velocity = 55km/h
3.) Acceleration = 94.29 km/h^2
Explanation:
Given that the train takes 35 minutes to cover the distance between the two towns. The train's average speed is 55 kilometers per hour. That is,
Time t = 35 minute
Convert it to hours by dividing it by 60
35/60 = 7/12 hours
Speed = 55 km/h
The displacement of the car is:
Displacement = 55 × 7/12 = 32.08 m
The velocity of the car will be 55 km/h
The acceleration of the car will be:
Acceleration = velocity/time
Substitute velocity and time into the formula
Acceleration = 55 ÷ 7/12
Acceleration = 94.29 km/h^2