Answer:
0.64 J/g°C
Explanation:
Using the formula;
Q = m × c × ∆T
Where;
Q = amount of heat
m = mass (g)
c = specific heat capacity
∆T = change in temperature (°C)
In this case:
Q (water) = - Q (metal)
mc∆T (water) = - mc∆T (metal)
According to the information in this question,
For water; m = 100g, c = 4.18J/g°C, ∆T = (25°C - 20°C)
For metal; m = 50g, c =?, ∆T = (25°C - 90°C)
mc∆T (water) = - mc∆T (metal)
100 × 4.18 × (25°C - 20°C) = - {50 × c × (25°C - 90°C)}
100 × 4.18 × 5 = - {50 × c × -65}
2090 = -{-3250c}
2090 = 3250c
c = 2090/3250
c = 0.643
c = 0.64J/g°C
Answer:
a) 0.00996 m
b) 109090909 Pa
Explanation:
Unit conversions:
1.2 mm = 0.0012 m
8.5 kN = 8500 N
If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is
With elastic modulus being E = 200 GPa, then its maximum stress must be
Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area
And its corresponding diameter is
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
<u>τ = 4900 N.m</u>
<u></u>
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
<u>τ = 4900 N.m</u>
<u></u>
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
<u>τ = 4900 N.m</u>
<u></u>
Hence, the correct answer will be:
<u>e. The torque is the same for all cases.</u>
Answer:
3.99*10^-3N/C
Explanation:
Using
Ep= kq/r²
Where r = 0.6mm = 0.6*10^-3m
K= 8.9*10^9 and q= 1.6*10^-19
So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²
= 3.99*10^-3N/C
What is it asking? It is sort of blurry
