The change in momentum of Sarah is equal to the impulse, which is the product between the force exerted by the seatbelt on Sarah and the time during which the force is applied:

$\Delta p=I\\m \Delta v = F \Delta t$

where

m is the mass

$\Delta v$ is the change in velocity

F is the average force

$\Delta t$ is the duration of the collision

In this problem:, we have:

m = 55 kg is Sarah's mass

$\Delta v = 0-20 = -20 m/s$ is the change in velocity

$\Delta t = 0.5 s$ is the duration of the collision

Solving for F, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

Where the negative sign indicates that the direction of the force is opposite to that of Sarah's initial velocity.

5 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

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

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

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

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

An air conditioner operating at steady state maintains a dwelling at 70oF on a day when the outside temperature is 90oF. If the

torisob [31]

30000 btuh /3413 btuh/kw. = 8.8 kw

8.8 kw/.746 kw/hp = 11.8 hp if COP is 1

11.8/3 hp (COP coefficient of performance) = 3.99 COP

>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.

3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.

8 0

2 years ago