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Sonja [21]
3 years ago
15

A person lifts a 4.5 kg cement block a vertical distance of 1.2 m and then carries the block horizontal- ly a distance of 7.3 m.

determine the work done by the person and by the force of gravity in this process.
Physics
1 answer:
Delvig [45]3 years ago
6 0
<span>When a person lifts the block, the block has more potential energy. Therefore the person does positive work on the block. work = m g h work = (4.5 kg) (9.80 m/s^2) (1.2 m) work = 52.92 joules The person's work on the block is 52.92 joules When the block is being raised, the force of gravity opposes the motion. Therefore the force of gravity does negative work on the block. work = - (force) (h) work = - m g h work = -(4.5 kg) (9.80 m/s^2) (1.2 m) work = -52.92 joules The work done by the force of gravity on the block is -52.92 joules Note that when the block is moved horizontally, the potential energy does not change. Therefore there is no work done on the block when it moves horizontally (we are assuming that the kinetic energy does not change).</span>
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A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the
Gelneren [198K]

Answer:

Option B. 2.8 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 27 m/s

Angle of projection (θ) = 30

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = 2 × 27 × Sine 30 / 9.8

T = 2 × 27 × 0.5 / 9.8

T = 27 / 9.8

T = 2.8 s

Therefore, time of flight of the ball is 2.8 s

7 0
3 years ago
A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down
KIM [24]
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
   EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
   EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
   Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
6 0
3 years ago
How did planck find the correct curve for the specturm of light emitted by a hot obkect?
Neko [114]

Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.

<h3>Briefing :</h3>
  • The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied. 
  • This is Planck's renowned equation for a black body's energy density.
  • According to this, electromagnetic radiation from heated bodies emits in discrete energy units or quanta, the size of which depends on a fundamental physical constant (Planck's constant). The basis of infrared imaging is the correlation between spectral emissivity, temperature, and radiant energy, which is made possible by Planck's equation.

Learn more about the Planck's constant with the help of the given link:

brainly.com/question/27389304

#SPJ4

3 0
1 year ago
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

4 0
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madreJ [45]
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