By using the formula
Force =mass×acceleration
Force=45×0.85
=38.25N
<span>Rising or falling, it does not change.</span>
Answer:
50m [N]
Explanation:
Think of these directions as if they were on a 2D plane. (if you're talking about displacement)
If you go 32m [N] then 6m [S], it's like you're moving backwards (-). (thus, 32n-6s=26m [N])
Next, you go north again, so moving forwards (+). (thus, 26n+24n=50m [N])
From the basic "heat lost by hot object=heat gained by colder object" principle, we have
m1c1ΔT1=m2c2ΔT2
where m1= 1kg
m2=4kg
c1=900J/kg k
c2=4200J/kg k
With this information at hand we have
m1c1(90-T)=m2c2(T-25)
after substituting the given values we can find that
T=28.3^{0}c
Answer:
6) 2.6 m/s, 31°
7) 9.2 m/s
8) 1.2 s
Explanation:
I'll do #6, #7, and #8 as examples. You can solve #9 using the equation from #7, and #10 using the equation from #8.
6) Take north to be +y and east to be +x.
Given:
vₓ = 2.2 m/s
vᵧ = 1.3 m/s
Find: v
v² = vₓ² + vᵧ²
v² = (2.2 m/s)² + (1.3 m/s)²
v ≈ 2.6 m/s
θ = atan(vᵧ / vₓ)
θ = atan(1.3 / 2.2)
θ ≈ 31°
7) Given:
Δy = -4.3 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)
v ≈ 9.2 m/s
8) Given:
Δy = -6.7 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.2 s