The period of a 261 hertz sound wave is
.000383 seconds
Answer:
6.96 s
Explanation:
<u>Given:</u>
- u = initial speed of the automobile = 0 m/s
- a = constant acceleration of the automobile =
- v = constant speed of the truck = 8.7 m/s
<u>Assume:</u>
- t = time instant at which the automobile overtakes the truck.
At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.
Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
741 J/kg°C
Explanation:
Given that
Initial temperature of glass, T(g) = 72° C
Specific heat capacity of glass, c(g) = 840 J/kg°C
Temperature of liquid, T(l)= 40° C
Final temperature, T(2) = 57° C
Specific heat capacity of the liquid, c(l) = ?
Using the relation
Heat gained by the liquid = Heat lost by the glass
m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)
Since their mass are the same, then
C(l)ΔT(l) = C(g)ΔT(g)
C(l) = C(g)ΔT(g) / ΔT(l)
C(l) = 840 * (72 - 57) / (57 - 40)
C(l) = 12600 / 17
C(l) = 741 J/kg°C
