Answer:
a)-2m/s^2
b)27.2m/s
Explanation:
Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)
W=455N=weight
W=mg
W=455N=weight
The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration
for point b we use the equations of motion with constant acceleration to find the velocity
Where
Vf = final speed
Vo = Initial speed
=0
A = acceleration
=2m/s
X = displacement
=6.8m
Solving
Can you show the diagram please
Answer:
Here is the complete question:
https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632
a) Current for long straight wire
b) Current at the center of the circular coil
c) Current near the center of a solenoid
Explanation:
⇒ Magnetic Field due to long straight wire is given by (B),where
Plugging the values,
Conversion ,and
⇒Magnetic Field at the center due to circular coil (at center) is given by,
So
⇒Magnetic field due to the long solenoid,
Then
So the value of current are , and respectively.
Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:
where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:
where in this problem:
is the initial pressure of the gas
is the initial absolute temperature of the gas
is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:
The potential energy of an object is defined by the equation: PE = mgh, where m = the mass of the object, g = the gravitational acceleration and h = the object's height above the ground.