All categories

3 years ago

A snail can move approximately 0.30 meters per minute.how many meters can the snail cover in 15 minutes?

Physics

1 answer:

ololo11 [35]3 years ago

3 0

Answer:

4.5 meters is your Answer goood luck please give 5 star

Explanation:

You might be interested in

A copper transmission cable 170 km long and 10.0 cm in diameter carries a current of 100 A .

Bond [772]

Answer:

a = = 37.2V

b = 13.39MJ

Explanation:

Given that

L = 170 × 10³

r = d/2

= 10cm / 2 = 5 cm

current I = 100A

we are to find the potential drop across the cable

so, we can use ohm’s law

V = IR = I (ρL/A)

ρ = resistivity of the copper

= 1.72 × 10⁻⁸ Ω.m

A = πr²

V = I(ρL/πr²)

= 100 (( 1.72 × 10⁻⁸ * 170 × 10³) / ( π * 0.05²))

= 37.2V

(b)

Energy (loss) = Pt

Enery (loss) = IVt

3600s per hour

= (100A)(37.2V)(3600s)

= 13.39MJ

8 0

4 years ago

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

erastovalidia [21]

Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

P₁V₁/T₁ = P₂V₂/T₂

1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

Final volume ≈ 231 cm³

7 0

3 years ago

Read 2 more answers

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons

Ostrovityanka [42]

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron

$v = \dfrac{I}{neA}$

$v = \dfrac{143}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 40.2 \times 10^{-6}}$

v = 2.615 × 10⁻⁴ m/s

time require is

$t=\dfrac{L}{v}$

$t=\dfrac{0.845}{2.615\times 10^{-4}}$

t = 3231.35 s

5 0

4 years ago

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.

lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

$\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\$

Δtm is time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

$\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr$

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0

3 years ago

A person hears the echo of his own voice from a distant hill after 2 seconds. How far away is the person from the hill, if the s

olga nikolaevna [1]

Answer: 330 m

Explanation:

The speed of sound $V$ is defined as the distance traveled by the sound wave $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=330 m/s$ is the speed of sound

$t=\frac{2 s}{2}=1 s$ is half the time the sound wave travels since the person speaks, the sound wave hits the hill and then returns to the person again as an echo

$d$ is the distance between the person and the hill

So, we have to find $d$ :

$d=Vt$

$d=(330 m/s)(1 s)$

Finally:

$d=330 m$

4 0

3 years ago