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2 years ago

Together, two students exert a force of 825 N in pushing a car a distance of 35 m.

Physics

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

28875 Joule

Explanation:

W = F * d

W = 825*35 = 28875 J

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PLEASEEEEE HELP ME !!!

Elza [17]

Answer:

maybe its heat sorry if it's wrong

because if friction is not in the problem so we are making heat or thermal energy

4 0

2 years ago

A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the

Alekssandra [29.7K]

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

$time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m$

Therefore, the distance between the submarine and the ocean floor is 11,250 m

3 0

2 years ago

Which of the following is a transformer used for?

mamaluj [8]

A transformer is used to increase or decrease a voltage. BUT ... it has to be an AC voltage, or the transformer doesn't work.

5 0

3 years ago

Read 2 more answers

it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f

Dahasolnce [82]

Answer:

$9.93\times 10^{19}Hz$

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

$f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz$

Therefore, the frequency equals to $9.93\times 10^{19}Hz$

4 0

2 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒ 0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0 and x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m

y = ? when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒ y = 18.4 m

7 0

3 years ago

Read 2 more answers