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just olya [345]
3 years ago
5

An athlete swims the length of 50m pool in 20sec, and

Physics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

1. 2.5 m/s

2. 2.27 m/s

3. 2.38 m/s

Explanation:

easy pz

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The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by th
Dmitry [639]

Answer:

d = V/E

Explanation:

From the definition, we can say that the electric field strength between the plates of a parallel plate capacitor is

E = v/d

where

E = electric field strength

V = potential difference

d = distance between the plates

On rearranging the equation and making d subject of the formula, we have

d = V/E

From the question, we're given that

V = 112 V

E = 1.12 kV/cm converting to V/m, we have 110000 V/cm

d = 112 / 110000

d = 0.00102 m

d = 1.02*10^-3 m

5 0
3 years ago
Determine the potential difference between two charged parallel plates that are 0.20 cm apart and have an electric field strengt
Nesterboy [21]
V=Ed=0.20\times 6=1.20
8 0
2 years ago
Read 2 more answers
Who first used the word atom to describe the smallest unit
Firlakuza [10]

Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.

Explanation: hope this helped

4 0
3 years ago
What coefficients would balance the following equation? __C2H6 + __O2 → __CO2 + __H2O A. 1C2H6 + 5O2 → 2CO2 + 3H2O B. 2C2H6 + 5O
sladkih [1.3K]
It's C, with the 2/7/4/6 in front of each reactant and product.
4 0
3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0
2 years ago
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