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3 years ago

10

Please help me with physics a very nice person

1 answer:

enot [183]3 years ago

3 0

14.8 g/cm^3 is the density

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3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0

3 years ago

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?(b) What is its wavelength in glass t

inn [45]

(a) $4.74 \cdot 10^{14}Hz$

The frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where

v is the wave's speed

$\lambda$ is the wavelength

For the red laser light in this problem, we have

$v=c=3\cdot 10^8 m/s$ (speed of light)

$\lambda=632.8 nm=632.8\cdot 10^{-9} m$

Substituting,

$f=\frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m}=4.74 \cdot 10^{14}Hz$

(b) 427.6 nm

The wavelength of the wave in the glass is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0 = 632.8\cdot 10^{-9} m$ is the original wavelength of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$\lambda=\frac{632.8\cdot 10^{-9}m}{1.48}=427.6\cdot 10^{-9}m=427.6 nm$

(c) $2.02\cdot 10^8 m/s$

The speed of the wave in the glass is given by

$v=\frac{c}{n}$

where

$c = 3\cdot 10^8 m/s$ is the original speed of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.02\cdot 10^8 m/s$

5 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

$D$ = distance between sue and van = 160 m

$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

hence there will be collision

7 0

3 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

3 years ago