What force would be required to accelerate a 1.100 kg car to 0,5 m/s27

The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>

Mass of block 1 = m1
Mass of block 2, = m2
Height of block 1 above the ground, = h1
Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

5 0

2 years ago

What is the acceleration of a car that increases its velocity from 0 to 100 kilometers per hour in 10 seconds?

Lunna [17]

Answer:

0.00278 km/sec2 or 36,000 km/hr2 .

Explanation:

5 0

2 years ago

An acorn falls from a tree. Its velocity just before it hits the ground is 28.2 m/s, downward. (acceleration of gravity is 9.81m

ss7ja [257]

Answer:

12.74 ms^-1 download

Explanation:

v=28.2, a=9.81

start from rest u=0

v=u+at=0+(9.81)t=28.2

t=2.875...

it reach 1.4 second before hitting the ground:

t=1.4, u=0, a=9.81

v=u+at=0+(9.81)(1.4)=12.74

7 0

2 years ago

What is Physics? How is Physics used in your everyday life? Give at least two examples and explain your answer. You need at leas

ikadub [295]

Answers:

The transportation industry is no stranger to the manipulation of everyday physics. Cars and trains utilize the wheel, which provides a smooth, steady motion.

The ears hear sounds which occur through the movement of air molecules, and the chemistry that drives all of biology depends on the physics of energy and molecules. Every day, for example, plants absorb sunlight, water, and carbon dioxide, creating glucose and releasing oxygen as a byproduct.

Brainlist pls!

3 0

2 years ago

Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.

iren [92.7K]

Answer:

F_n = 5.65E-11 N

d = 1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

$F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}$

For $F_n$

$F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}$

Dividing the above equations we get

$\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N$

F_n = 5.65E-11 N

$F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m$

d = 1.20682E-31 m

8 0

3 years ago