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I am Lyosha [343]
3 years ago
12

PLEASE HELP - basic SUVAT stuff

Physics
1 answer:
Goryan [66]3 years ago
6 0

Answer:

1. 5 m/s

2. 8 m/s

3. 10 s

4. 3 m/s

Explanation:

1. Average speed = distance / time

s = 20 m / 4 s

s = 5 m/s

2. First, find the time to travel from A to B:

60 m = 10 m/s × t

t = 6 s

Now find the average speed from A to C:

s = 80 m / (6 s + 4 s)

s = 8 m/s

3. If t is the time traveling from C back to B, and the average speed of the whole trip is 5 m/s, then:

5 m/s = 100 m / (6 s + 4 s + t)

t = 10 s

4. Average velocity = displacement / time

v = 60 m / 20 s

v = 3 m/s

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It would be 300 cm/h
and 5 cm/m
4 0
3 years ago
A wave is produced in a rope. The wave has a speed of 33 m/s and a frequency of 22 Hz. What wavelength is produced? 0. 67 m 0. 7
vekshin1

Answer:

Wavelength = <u>1.5 m</u>

Explanation:

The formula for waves in terms of wavelength, speed and frequency is:

Speed (v) = Frequency (f) × Wavelength (λ)

33 = 22 × λ

33 = 22λ

λ = \frac{33}{22}

So, λ = 1.5 m

4 0
2 years ago
13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i
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Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0
2 years ago
A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

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\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

7 0
3 years ago
The atomic mass of an element is
inessss [21]

Here are the answers to the question. Make sure to give a valid reason, please.

A. the sum of the protons and neutrons in one atom of the element.

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C. a weighted average of the masses of an element's isotopes.

D. twice the number of protons in one atom of the element.

6 0
3 years ago
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