Answer:
The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2
Explanation:
Given:
F = 550 N
m = 1000 kg
To Find:
a = ?
Solution:
So by the equation by Newton's 2nd Law of Motion,
F = m x a
550 N = 1000 kg x a
a = 550 N/ 1000 kg
a = 0.55 m/s^2
Therefore,
The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2
PLEASE MARK ME AS BRAINLIEST!!!
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.
At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.
The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.
-- If the object is falling downward, it moves 9.8 m/s faster every second.
-- If the object is tossed upward, it moves 9.8 m/s slower every second.
The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .
(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)
Answer:
For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.
Explanation:
For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.
When a liquid is cooled, the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases, and the matter changes its state to solid.
Answer:
I would love to help, Could you put the question in English?
Explanation:
Explanation:
It is given that,
Velocity of the electron, ![v=(2\times 10^6i+3\times 10^6j)\ m/s](https://tex.z-dn.net/?f=v%3D%282%5Ctimes%2010%5E6i%2B3%5Ctimes%2010%5E6j%29%5C%20m%2Fs)
Magnetic field, ![B=(0.030i-0.15j)\ T](https://tex.z-dn.net/?f=B%3D%280.030i-0.15j%29%5C%20T)
Charge of electron, ![q_e=-1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q_e%3D-1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
(a) Let
is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :
![F_e=q_e(v\times B)](https://tex.z-dn.net/?f=F_e%3Dq_e%28v%5Ctimes%20B%29)
![F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]](https://tex.z-dn.net/?f=F_e%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20%5B%282%5Ctimes%2010%5E6i%2B3%5Ctimes%2010%5E6j%29%5Ctimes%20%280.030i-0.15j%29%5D)
![F_e=-1.6\times 10^{-19}\times (-390000)(k)](https://tex.z-dn.net/?f=F_e%3D-1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20%28-390000%29%28k%29)
![F_e=6.24\times 10^{-14}k\ N](https://tex.z-dn.net/?f=F_e%3D6.24%5Ctimes%2010%5E%7B-14%7Dk%5C%20N)
(b) The charge of electron, ![q_p=1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q_p%3D1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.