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Paha777 [63]
3 years ago
14

What is environmental cleanliness​

Physics
1 answer:
maw [93]3 years ago
8 0

Answer:

it's the condition of the Earth to become neat and organized. it will give negative impact on the society and teach people to clean to have a better world to stay

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A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs
Maksim231197 [3]

Answer:

None, both objects will hit ground at the same time.

Explanation:

  • Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

        \Delta h = \frac{1}{2}*g*t^{2}  (1)

  • As the left side in (1) is the same for both objects, the right side will be the same also.
  • Since g is constant close to the surface of the Earth, it's also the same for both objects.
  • So, the time t must be the same for both objects also.
6 0
3 years ago
A block (6 kg) initially compresses spring #1 (k = 2000 N/m) by 60 cm from its equilibrium point. When the block is released, it
mart [117]

Answer:

block K = 29.39 J and spring #1   Ke = 360 J

Explanation:

In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work

        W_{fr}= Ef - E₀

Let's look for the energies

Initial

        E₀ = Ke = ½ k₁  x₁²

Final, this is just before starting to compress the spring

        Ef = Ke = ½ m v²

The work of the rubbing force is

       W_{fr}= -fr x

Let's write Newton's second law the y axis

       N-W = 0

      N = W

      fr = μ N

      fr = μ mg

Let's replace

      -μ mg x = ½ m v² - ½ k₁ x₁²

       v² = 2/m (½ k₁ x1₁² -μ mg x)

       v² = 2/6  (½ 2000 0.6²2 - 0.5  6  9.8 1) = 1/3 (360 - 29.4)

       v = 3.13 m / s

With this value we calculate the energy of the block

       K = ½ m v²

       K = ½  6  3.13²

       K = 29.39 J

Calculate eenrgy of the spring ke 1

      Ke = ½ k₁ x₁²

      Ke = ½ 2000 0.60²

      Ke = 360 J

4 0
3 years ago
Scientific experiments must be able to be repeated by multiple scientists to verify the results that are obtained. Which of the
Murrr4er [49]
A. Why do people like to watch TV
7 0
3 years ago
Sarah is demonstrating the gravitational force on falling objects to her class. she drops an 11 lb. bowling ball from the top of
GalinKa [24]
C. 29 m/s

Basing on the information given, we can compute for the velocity with the following
Mass = 11 lb. convert to kg = 5 kg
Gravity = 9.8 m/s2
Time (final) = 3.0 s

v= at
v =  9.8m/s2(3.0s) 
= 29.4 m/s
Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
4 0
3 years ago
The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R
Andrew [12]

Answer:

The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

Explanation:

Given that,

Voltage = 64.5

Resistance is

R_{1}=711\ \Omega

R_{2}=182\ \Omega

R_{3}=663\ \Omega

R_{4}=534\ \Omega

R_{5}=265\ \Omega

Suppose, the specified points are R₁ and H.

According to figure,

R₂,R₃,R₄ and R₅ are connected in parallel

We need to calculate the resistance

Using parallel formula

\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}

Put the value into the formula

\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}

\dfrac{1}{R}=\dfrac{35501}{2806615}

R=79.05\ \Omega

R and R₁ are connected in series

We need to calculate the equilibrium resistance

Using series formula

R_{eq}=R_{1}+R

R_{eq}=711+79.05

R_{eq}=790.05\ \Omega

We need to calculate the equivalent current

Using ohm's law

i_{eq}=\dfrac{V}{R_{eq}}

Put the value into the formula

i_{eq}=\dfrac{64.5}{790.05}

i_{eq}=0.0816\ A

We know that,

In series combination current distribution in each resistor will be same.

So, Current in R and R₁ will be equal to i_{eq}

The current at h point will be equal to current in R₅

We need to calculate the voltage in R

Using ohm's law

V=I_{eq}\timesR

Put the value into the formula

V=0.0816\times79.05

V=6.45\ Volt

In resistors parallel combination voltage distribution in each part will be same.

So, V_{2}=V_{3}=V_{4}=V_{5}=6.45 V

We need to calculate the current at H point

Using ohm's law

i_{h}=\dfrac{V_{5}}{R_{5}}

Put the value into the formula

i_{h}=\dfrac{6.45}{265}

i_{h}=0.0243\ A

Hence, The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

5 0
3 years ago
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