Answer:
451.13 J/kg.°C
Explanation:
Applying,
Q = cm(t₂-t₁)............... Equation 1
Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.
Make c the subject of the equation
c = Q/m(t₂-t₁).............. Equation 2
From the question,
Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C
Substitute these values into equation 2
c = 1500/[0.133(45-20)]
c = 1500/(0.133×25)
c = 1500/3.325
c = 451.13 J/kg.°C
Answer:
2.78 m
Explanation:
At the peak, the velocity is 0.
Given:
a = -1.6 m/s²
v₀ = 2.98 m/s
v = 0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)
x = 2.775 m
Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.
Average speed = (total distance covered) / (time to cover the distance)
Total distance = (80m) + (125m) + (45m) = 250 meters
Overall time = 10 minutes
Average speed = (250 meters) / (10 minutes)
<em>Average speed = 25 meters/minute </em>
Since we're only looking for average speed and not velocity, we don't care about any of the directions, and we don't need to calculate Mary's displacement.
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm
Answer is B 3000 J hope I could help
