Answer:
Q = C M T where C is the specific, M the mass, T the temperature change
Note 1 cal = 4.19 Joules
1562.75 J / (4.19 J/cal) = 378 cal
C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)
C = .096 cal / g deg C
Answer:
DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER
Explanation:
Assuming the liquid to be water ,
the density
of water is :
Buoyant force exerted by a liquid on an object with
of it's volume immersed is :

where ,
is the buoyant force
is the density of the liquid
is the acceleration due to gravity
Thus at equilibrium:

from these , we get the density of brass to be 
which is not possible
<h2>Given that,</h2>
Mass of two bumper cars, m₁ = m₂ = 125 kg
Initial speed of car X is, u₁ = 10 m/s
Initial speed of car Z is, u₂ = -12 m/s
Final speed of car Z, v₂ = 10 m/s
We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

v₁ is the final speed of car X.

So, car X will move with a velocity of -12 m/s.
The acceleration of the car would be 0.33 first and then it would be 0.17.
<u>Explanation:</u>
An applied force is a force that is applied to an object by an individual or another item. On the off chance that an individual is pushing a work area over the room, at that point there is an applied power following up on the article. The applied power is the power applied on the work area by the individual.
The net force applied to the object rises to the mass of the article increased by the measure of its acceleration. The net power following up on the soccer ball is equivalent to the mass of the soccer ball duplicated by its adjustment in speed each second (its acceleration).
Answer:
Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.
Explanation:
The magnetic field, B of long straight wire can be obtained by applying ampere's law

I is here current, and r's the distance from the wire to the field of measurement.
The magnetic field is obviously directly proportional to the current wire. From this expression.
As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.