The process of splitting one large nucleus into
smaller ones is nuclear fission.
The process of combining two small nuclei into
one larger one is nuclear fusion.
Answer:
672000 J.
Explanation:
From the question,
q = cm(t₂-t₁)............................ Equation 1
Where q = heat energy supplied to the water, c = specific heat capacity of water, m = mass of water, t₂ = final temperature, t₁ = Initial temperature.
Given: m = 8.0 kg, c = 4200 J/kg°C, t₂ = 40°C, t₁ = 20°C
Substitute these values into equation 1
q = 8×4200×(40-20)
q = 8×4200×20
q = 672000 J.
Hence the energy suplied to the water is 672000 J.
Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
Answer:
B.useful products
Explanation:
industry is a sector that produces goods or services within an economy
Answer:
The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Explanation:
Given;
the speed of the slower runner, u₁ = 11.8 km/hr
the speed of the fastest runner, u₂ = 15 km/hr
distance, d = 8 km
The time when the fastest runner finishes the race is given by;
The distance covered by the slower runner at this time is given by;
d₁ = u₁ x 0.533 hr
d₁ = 11.8 km/hr x 0.533 hr
d₁ = 6.29 km
Additional distance (x) the slower runner need to finish is given by;
6.29 km + x = 8km
x = 8 k m - 6.29 km
x = 1.71 km
Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.