Answer:
0.24
Explanation:
Mass of ball= 12g=0.012Kg
height of ball= 2.5m
velocity of ball before falling= 3.2m/s
potential energy of the ball=mgh= 0.012*10*2.5=0.3J
kinetic energy of the ball=0.5*m
=0.5*0.012*3.2*3.2=0.6J
Loss in mechanical energy during the fall= potential energy- Kinetic energy= 0.3-0.06=0.24J
note: During the fall, the potential energy of the ball is converted to kinetic energy. the loss in energy is due to air resistance.
Answer:
1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity
2) An upward force of magnitude 5 N is exerted on the book by the table
Explanation:
First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.
For the book in the problem, the magnitude of the weight is 5 N.
We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.
Below the crust is the mantle, a dense, hot layer of semi-solid rock approximately 2,900 km thick. The mantle, which contains more iron, magnesium, and calcium than the crust, is hotter and denser because temperature and pressure inside the Earth increase with depth.
Hope this helped! :)
Can I be brainliest please?
The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore
The vector b is 41.4° up from the x-axis. Therefore
![\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bb%7D%20%3D%20b%5Bcos%2841.4%5E%7Bo%7D%29%20%5Chat%7Bi%7D%20%2B%20sin%2841.4%5E%7Bo%7D%29%20%5Chat%7Bj%7D%20%5D%20%3Db%280.75%5Chat%7Bi%7D%20%2B%200.6613%20%5Chat%7Bj%7D%29)
The vector a is 27.7° up from the x-axis. Therefore
![\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%20%3D%20a%5Bcos%2822.7%5E%7Bo%7D%29%5Chat%7Bi%7D%20%2B%20sin%2827.7%5E%7Bo%7D%29%5Chat%7Bj%7D%5D%20%3D%20%20a%280.8854%5Chat%7Bi%7D%20%2B%200.4648%5Chat%7Bj%7D%29)
Because
, the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)
Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854
Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m
To solve the problem, we
must know the heat capacity of ice and water.
For Cp = 2090 J/kg C
H = mCpT
H = (10 kg) ( 2090 J/ Kg C)
( -23 C)
H = - 480700 J
For water Cp = 4180 j/kg C
H = (100 kg) ( 4180 J/kg C)
( 60 C)
<span>H = 2508000 J</span>
