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OverLord2011 [107]
2 years ago
7

Where does reduction occur in an electrolytic cell

Physics
1 answer:
den301095 [7]2 years ago
3 0

Answer:

The answer is actually a. Cathode

Explanation:

Just took the quiz

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You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30°C to
earnstyle [38]

The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.

We are asked to calculate the velocity of sound at 36 degree celsius.

Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.

The velocity at this temperature is calculated as -

                            V = 331  + 0.6T m/s

                               = 331 + 0.6 × 36 m/s

                               = 331 + 21.6 m/s

                               = 352.6 m/s.

Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

5 0
3 years ago
Read 2 more answers
2. A cinder block is sitting on a platform 20 m high. It weighs 16kg. The block has
Cerrena [4.2K]

Answer:

3136 Joules

Explanation:

Applying,

P.E = mgh.............. Equation 1

Where P.E = potential energy, m = mass of the cinder block, h = height of the platform, g = acceleration due to gravity.

From the question,

Given: m = 16 kg, h = 20 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 16(20)(9.8)

P.E = 3136 Joules

Hence the potential energy of the cinder block is 3136 Joules

7 0
2 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
You’re driving down the highway late one night at 20 m/s when a deer steps out onto the road 35 m in front on you. Your reaction
Salsk061 [2.6K]

Answer:

i would juh hit da deer no kap

Explanation:

4 0
3 years ago
Give examples of not useful high friction
Tatiana [17]
Friction can be bad by being too strong or too weak. 

<span>Sometimes, when it is too strong, it decreases efficiency since some energy is wasted and turns to heat. Friction can also d</span><span>amage equipment/objects like when you slide it on the floor.
</span>
When friction is too weak, like for instance when there is black ice- our center of gravity is displaced too quickly and we can fall. Likewise, if there is a lot of slush on the ground, cars can slip and slide.
6 0
3 years ago
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