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OLga [1]
3 years ago
5

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

.
Physics
2 answers:
Bogdan [553]3 years ago
8 0

Answer:

m = 40.8 [kg]

Explanation:

We know that weight is defined as the product of mass by gravitational acceleration in this way we have the following equation.

W = m*g

where:

W = weight [N] (units of Newtons)

m = mass [kg]

g = gravity acceleration = 9.8 [N/kg]

Therefore:

m=W/g\\m=400/9.8\\m=40.8[kg]

Sonbull [250]3 years ago
8 0

Answer: 41!

Explanation:

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In the first video for chapter 29, we looked at the emission spectrum of excited gases. In this video, we look at the __________
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Although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.

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2 years ago
Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe
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Sarah's acceleration is -0.49 m/s^2

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

F_f = \mu mg

where

\mu is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

F=ma

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

F=-\mu mg = ma

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

a=-\mu g

And substituting the following values:

\mu = 0.05 (coefficient of friction)

g=9.81 m/s^2 (acceleration of gravity)

we find:

a=-(0.05)(9.81)=-0.49 m/s^2

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3 years ago
How many calendar days pass between the new moon and full moon?
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Read 2 more answers
A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel
marusya05 [52]

(a) 7.32\cdot 10^7 Hz

The frequency of an electromagnetic waves is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda=4.1 m is the wavelength of the wave in the problem

Substituting into the equation, we find

f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz

(b) 4.60\cdot 10^8 rad/s

The angular frequency of a wave is given by

\omega = 2\pi f

where

f is the frequency

For this wave,

f=7.32\cdot 10^7 Hz

So the angular frequency is

\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s

(c) 1.53 m^{-1}

The angular wave number of a wave is given by

k=\frac{2\pi}{\lambda}

where

\lambda is the wavelength of the wave

For this wave, we have

\lambda=4.1 m

so the angular wave number is

k=\frac{2\pi}{4.1 m}=1.53 m^{-1}

(d) 1.03\cdot 10^{-6}T

For an electromagnetic wave,

E=cB

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

I=\frac{E^2}{2\mu_0 c}

where we have

E = 310 V/m is the amplitude of the electric field

\mu_0 is the vacuum permeability

c is the speed of light

Substituting into the formula,

I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2

(g) 1.53\cdot 10^{-8} kg m/s

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

\frac{dp}{dt}=\frac{A}{c}

where the <S> is the magnitude of the Poynting vector, given by

=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2

and where the surface is

A = 1.8 m^2

Substituting, we find

\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s

(h) 8.47\cdot 10^{-7} N/m^2

For a surface that totally absorbs the wave, the radiation pressure is given by

p=\frac{}{c}

where we have

=254.2 W/m^2

c=3\cdot 10^8 m/s

Substituting, we find

p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2

8 0
3 years ago
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