Heat used by electric heater :
Q = m • c • ∆T
Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)
Q = 8.82 × 10⁶ J
Cost of electrical energy :
Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)
Cost = $ 0.3675
First method
initial distance = 16m
final distance= 43 m
total distance covered= final -initial
=43m -16m
=27m
Second method
Si= 16m
Sf =43 m
t= 12 s
first we will find V
V = (Sf-Si)/ t
V =( 43- 16)/ 12
V = 27/12 ⇒ V= 9/4
V= distance / time
distance= V×time
distance = (9/4) ×12
distance =27
Answer
Given,
Average speed of Malcolm and Ravi = 260 km/h
Let speed of the Malcolm be X and speed of the Ravi Y.
From the given statement
....(i)
....(ii)
Adding both the equations
3 X = 600
X = 200 km/h
Putting value in equation (i)
Y = 520 - 200
Y = 320 Km/h
Speed of Malcolm = 200 Km/h
Speed of Ravi = 320 Km/h
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
Answer:
371.2 mm
Explanation:
The Balmer series of spectral lines is obtained from the formula
1/λ = R(1/2² -1/n²) where λ = wavelength, R = Rydberg's constant = 1.097 × 10⁷ m⁻¹
when n = 15
1/λ = 1.097 × 10⁷ m⁻¹(1/2² -1/15²)
= 1.097 × 10⁷ m⁻¹(1/4 -1/225)
= 1.097 × 10⁷ m⁻¹(0.25 - 0.0044)
= 1.097 × 10⁷ m⁻¹ 0.245556
= 2.693 10⁶ m⁻¹
So,
λ = 1/2.693 10⁶ m⁻¹
= 0.3712 10⁻⁶ m
= 371.2 mm
