Answer:
b. 600,000 J
Explanation:
Applying the law of conservation of energy,
The thermal energy created = Kinetic energy of the suv.
Q' = 1/2(mv²)............... Equation 1
Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.
From the question,
Given: m = 3000 kg, v = 20 m/s
Substitute these values into equation 1
Q' = 1/2(3000×20²)
Q' = 600000 J
Hence the right option is b. 600,000 J
Answer:
The current through the resistor is 0.5 A
Explanation:
Given;
power of the light bulb = 60 W
voltage in the wall outlet across the plug terminals = 120 V
power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.
power = voltage x current

Therefore, for a 60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A
Answer:
A morse code alphabet decoder maybe?? I am confused
Explanation:
Answer:
A. 24 m, 14 m/s
B. 8.0 m
Explanation:
Given:
x₀ = 6.0 m
v₀ = 4.0 m/s
a = 5.0 m/s²
t = 2.0 s
A. Find: x and v
x = x₀ + v₀ t + ½ at²
x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²
x = 24 m
v = at + v₀
v = (5.0 m/s²) (2.0 s) + (4.0 m/s)
v = 14 m/s
B. Find x when v = 6.0 m/s.
v² = v₀² + 2a (x − x₀)
(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)
x = 8.0 m
Answer:
a) T ’= 0.999 s
, b) t = 3596.4 s
Explanation:
The angular velocity of a simple pendulum is
w = √g / L
The angular velocity, frequency and period are related
w = 2π f = 2π / T
2π / T = √ g / L
T = 2π √ L / g
L = T² g / 4π²
L = 1² 9.8 / 4π²
L = 0.248 m
To know the effect of the temperature change let's use the thermal expansion ratios
ΔL = α L ΔT
ΔL = 24 10⁻⁶ 0.248 (-4 - 20)
ΔL = 142.8 10⁻⁶ m
Lf - L = -142. 8 10⁻⁶
Lf = 142.8 10⁻⁶ + 0.248
Lf = 0.2479 m
Let's calculate new period
T ’= 2π √ L / g
T ’= 2π √ (0.2479 / 9.8)
T ’= 0.999 s
We can see that the value of the period is reduced so that the clock is delayed
b) change of time in 1 hour
When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is
t = 3600 0.999
t = 3596.4 s
Therefore the clock is delayed almost 4 s