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ad-work [718]
3 years ago
9

How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?

Physics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

W = 47040  J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040  J

So, the required work done is 47040  J.

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table
MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

3 0
3 years ago
What chemicals do it take to make elephant toothpaste???
ladessa [460]

Answer:

you need water, sodium iodide, and soap

3 0
2 years ago
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Is a measure of how closely packed together the particles of matter are in a specific volume
Mnenie [13.5K]

Answer:

Density is an important physical property of matter. It reflects how closely packed the particles of matter are. When particles are packed together more tightly, matter has greater density.

Explanation:

5 0
3 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
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