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Dafna11 [192]
3 years ago
13

You wanted to use your electric grill in your backyard. You need an extension cord. What will be the requirement of your extensi

on cord in order to use your electric grill without any incidents. Base your explanation on " the resistance of a conducting wire increases with the increasing length of a wire".
Physics
2 answers:
Troyanec [42]3 years ago
8 0

Answer:

you need to be able to have long enough to reach and have it far away from things that are going to cause accidents

olasank [31]3 years ago
3 0
You need to be able to have long enough to reach and have it far away from things that are going to cause accidents
You might be interested in
An acorn with a mass of 0.300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.oo m/s?
vladimir1956 [14]

Answer: 3.75 joules

Explanation:

Given that:

Mass of acorn = 0.300 kilograms

velocity = 5.oo m/s

Kinetic energy = ?

Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 0.300kg x (5.00m/s)^2

= 0.5 x 0.3kg x (5.00m/s)^2

= 0.15 x (5.00m/s)^2

= 3.75 joules

Thus, the kinetic energy of the falling acorn is 3.75 joules

5 0
3 years ago
A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio
Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

 v² = u² + 2 a h

 0 = 64² - 2 x 32 x h

 64 h = 64²

  h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0
3 years ago
this stationary wave is what we call the first harmonic of the first normal mode of the system. in units of l, the length of the
pentagon [3]

First harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

<h3>First harmonic of a closed pipe</h3>

The first harmonic of a closed pipe is the fundamental frequency of the closed of the closed pipe.

L = λ/4

where;

  • L is the length of the pipe
  • λ is the wavelength of sound

λ = 4L

But, v = F₀λ

v = F₀(4L)

F₀ = v/4L

where;

  • F₀ is the first harmonic
  • v is speed of sound

Thus, first harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

Learn more about fundamental frequency here:  brainly.com/question/1967686

#SPJ11

<h3 />
6 0
1 year ago
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