It’s upside down, I know.

If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)

taurus [48]

Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20 Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories, 4.2 J = 1 Calorie
= 8 400 000 / 4.2 = 200 000

Q = 200 000 Calories

4 0

3 years ago

What force is needed to accelerate an object 5 m/s if the object has a mass of 10kg?

nevsk [136]

The <span>force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10</span>

6 0

3 years ago

The drawing below shows two different types of pulley systems designed to lift a weight. In pulley system A, the end of the rope

nadya68 [22]

The answer would be in the chart or graph A is 1 B is 2

4 0

3 years ago

Read 2 more answers

You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop

mina [271]

1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g
2. Find the acceleration due to friction:
F = m*a = μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t²
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s

7 0

3 years ago

Two forces F1 and F2 are acting on the box shown in the drawing, causing the box to move across the floor. The two force vectors

SpyIntel [72]

Answer: The correct answer is (b) F1 does more work than F2.

The block is moving horizontally the direction is opposite to direction of motion. The two forces applied are F1 and F2.F1 is applied horizontally so its horizontal component is F1cos0 = F1

While F2 is at angle so horizontal component of F2 is F2cos(theta) and as we know cos is a decreasing in 0 to 90 degree. Therefore F1 does more work than F2.

Work is defined as the product of the magnitude of a force and the displacement of the object it is acting upon. In this case, the forces F1 and F2 are acting on the box, causing it to move across the floor. To calculate the work done by each force, we need to calculate the magnitude of the force and the displacement of the box.

The magnitude of F1 is given in the diagram, and it is 2 N. The magnitude of F2 is also given, and it is 4 N. For the displacement of the box, we will assume it is 1 m.

Now we can use the formula W = F x d to calculate the work done by each force.

For F1: W1 = 2 N x 1 m = 2 J

For F2: W2 = 4 N x 1 m = 4 J

Since F1 does 2 J of work, and F2 does 4 J of work, F1 does more work than F2 does. Therefore, the correct answer is (b) F1 does more work than F2 does.

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7 0

1 year ago