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2 years ago

Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)

Physics

1 answer:

Mars2501 [29]2 years ago

3 0

Answer:

300 J

Explanation:

Given :

Applied force = 30 N
Mass = 2 kg
Height = 10 m
g = 10 m/s²

Work done by the force :

Work done = Force x Displacement
Work done = mg x h
Work done = 30 N x 10 m
Work done = 300 J

Note :

What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
But in the end, it asks for work done by the force of 30N
Hence, the given answer ~

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on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

DIA [1.3K]

Momentum
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v [solve for v]
v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.

6 0

3 years ago

Observe and compare the forces acting on the turtle and the cat.

Pepsi [2]

Answer:

The forces are balanced on both animals because they are not moving

More importantly than not moving is not accelerating.

Explanation:

5 0

2 years ago

Seafloor spreading is a process that occurs along ___________1____________. At ___________2____________ boundary, two plates mov

sergejj [24]

Answer:

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Explanation:

5 0

2 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

3 years ago

A train travels 79 kilometers in 2 hours, and then 77 kilometers in 3 hours. What is its average speed?

koban [17]

Average speed=total distance traveled/time. Total distance=the sum of the two short distances=79*2+77*3=389 km. Total time used to travel is 2+3=5 hr. Average speed is 389/5=77.8 km/hr.

4 0

3 years ago