Answer:
46°C
Explanation:
q = mCΔT
336,000 J = (20 kg) (4200 J/kg/°C) (50°C − T)
T = 46°C
Answer:
27 m
Explanation:
Given:
v₀ = 6 m/s
a = 2 m/s²
t = 3 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (6 m/s) (3 s) + ½ (2 m/s²) (3 s)²
Δx = 27 m
Answer:
a. 150 N
Explanation:
Gravitational Force: This is the force that act on a body under gravity.
The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.
The S.I unit of gravitational force is Newton(N).
Mathematically, gravitational force of attraction is expressed as
(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)
(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).
Given: m = 17 kg, g = 8.8 m/s²
Substituting into equation 2,
F = 17(8.8)
F = 149.6 N
F ≈ 150 N.
Thus the gravitational force = 150 N
The correct option is a. 150 N
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is

.
In fact, the laws of motions on both x- and y- directions are


From the second equation, we get the time t at which the projectile hits the ground, by requiring

, and we get:

And inserting this value into Sx(t), we find

And this value is maximum when

, so this is the angle at which the projectile reaches its maximum distance.
So now we can take again the law of motion on the x-axis

And by using

, we find the value of the initial velocity v0: