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2 years ago

Miles is camping in Glacier National Park. In the midst of a glacier canyon,

Physics

1 answer:

valentina_108 [34]2 years ago

8 0

Answer:

t=1.623 sec

Explanation:

The distance traveled before the echo is had is:

$distance=2d, d=280\ m\\\\=280\times 2\\\\=560 \ m$

Given the speed of sound as v=345m/s, we use the speed equation to solve for t:

$v=\frac{d}{t}\\\\345\ m/s=\frac{560m}{t}\\\\t=\frac{560}{360}\\\\=1.623 \ s$

Hence, it takes 1.623 seconds to hear the echo.

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A cylindrical tungsten filament 14.0 cmcm long with a diameter of 1.00 mmmm is to be used in a machine for which the temperature

Advocard [28]

Answer:

Resistivity ρ=1.12 x 10^-4 Ωm

Explanation:

ρ= RA/l, where R is resistance, A is cross sectional area and l is length

A=πr^2

Note Current is given R is proportion to temperature and inversely proportional to Current R=(20+273)/14*10^-2 =2000Ω

⇒ρ=R*πr^2/l all length in metre.

8 0

3 years ago

A car goes forward along a level road T a constant velocity the additional force needed to being the car into equilibrium is?

eimsori [14]

The additional force needed to bring the car into equilibrium is frictional force.

8 0

2 years ago

The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.

Assoli18 [71]

A magnetic flux would be the correct answer

8 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

4 0

3 years ago