<u>The Weight </u>is a vector whose magnitude is the product of the mass m of the object and the magnitude of the local gravitational acceleration. Its always directed toward the center of the Earth.
The density of the substance is obtained by dividing its mass by its volume. Density is an intensive property which means that its value does not depend on the amount of substance and will always stay the same for same conditions.
d = 84.7 g / 46.7 cm³ = 7.75 g / x cm³
The value of x is approximately 4.27 cm³.
Answer:
c=0.14J/gC
Explanation:
A.
2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.
B.
We can use the expression for heat transmission
In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say
for water we have to
c = 4.18J / g ° C
replacing we have
I hope this is useful for you
A.
2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.
B.
Podemos usar la expresión para la transmisión de calor
En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir
para el agua tenemos que
c=4.18J/g°C
reemplazando tenemos
Answer:
ω' = 0.815 rad/s
Explanation:
Given,
R = 1.20 m
Inertia of merry-go- round= 240 kg.m²
Rotating speed = 9 rpm =
=0.9424 rad/s
mass of the child, m = 26 kg
angular speed of the merry-go-round=?
we know
Angular momentum, L = I ω
Moment of inertia of the child
I' = m r² = 26 x 1.2² = 37.44 kgm²
Conservation of angular momentum
initial angular momentum = Final angular momentum
I ω = (I+I')ω'
240 x 0.9424 = (240+37.44) ω'
226.176= 277.44 ω'
ω' = 0.815 rad/s
new angular speed of the merry-go- round is equal to 0.815 rad/s
This would make it quantitative