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3 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

Physics

2 answers:

mel-nik [20]3 years ago

5 0

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

olasank [31]3 years ago

3 0

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

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4 years ago

Suppose a skydiver (mass =100kg) is falling towards the earth. When the skydiver is 80 m above the earth he is moving at 60 m/s

goblinko [34]

Answer:

The total mechanical energy of the skydiver is, E = 96402.6 J

Explanation:

Given data,

The mass of the skydiver, m = 100 kg

The speed of the skydiver at 80 m height, v = 60 m/s

The initial velocity of the skydiver, u = 0

Using the III equations of motion,

v² = u² + 2gs

s = v²/2g

Substituting the given values,

s = ½ 60²/ 9.8

= 18.37 m

Hence the initial total distance of the skydiver from the ground initially,

h = s + d

= 18.37 + 80

= 98.37 m

Since the total mechanical energy of a system is conserved, the total mechanical energy of the skydiver at height 'h' is equal to the total mechanical energy at height 'd'.

E = P.E + K.E

= mgh + ½ mu²

= 100 x 9.8 x 98.37 ( ∵ u = 0)

= 96402.6 J

Hence, the total mechanical energy of the skydiver is, E = 96402.6 J

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3 years ago

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4 years ago

A 25 V battery is connected into a circuit with two resistors in series. One resistor has resistance 5.0 O and the other has res

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3 years ago

Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc

Sphinxa [80]

Answer:

F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

V_{water} = 1.09 10⁻³ m³

the amount of water is

ρ = m / V

m = ρ V

m = 1000 1.09 10⁻³

m = 1.09 kg

so the weight of the liquid in the container for this time is

W = mg

W = 1.09 9.8

W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v = $\sqrt{0 -2 g ( 0-y_o)}$

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

P₁ = 1.28 10⁵ Pa

The definition of Pressure is

P = F / A

F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

F = P A

F = 1.28 10⁵ 7.46 10⁻⁵

F = 9.5 N

with these data let's use the equilibrium equation

F_ scale -W - F = 0

F_scale = W + F

F_scale = 10.682 + 9.5

F_scale = 20.18 N

4 0

3 years ago