The main explanation for why heat pumps work is the ________ law of thermodynamics. Zeroth First Second

Determine the magnitude and direction of the resultant velocity of 75.0 m/s. 25.0 east of north, and 100.0 m/s, 25.0 east of sou

aleksandr82 [10.1K]

Answer:

77.35 m / s

Ф = -17° from + X axis or 343° from + X axis

Explanation:

v1 = 75 m/s 25° east of north

v2 = 100 m/s 25° east of south

Write the velocities in vector form ,we get

$\overrightarrow{v_{1}}=75\left ( Sin25\widehat{i} +Cos25\widehat{j}\right )=31.7\widehat{i}+67.97\widehat{j}$

$\overrightarrow{v_{1}}=100\left ( Sin25\widehat{i} -Cos25\widehat{j}\right )=42.26\widehat{i}-90.63\widehat{j}$

Now add the velocity vectors to get the resultant of the velocities.

$\overrightarrow{v}=\overrightarrow{v_{1}}+\overrightarrow{v_{2}}$

$\overrightarrow{v}=\left (31.7+42.26 \right )\widehat{i}+\left ( 67.97- 90.63 \right )\widehat{j}$

$\overrightarrow{v}=73.96\widehat{i}-22.66\widehat{j}$

magnitude of resultant velocity is $\sqrt{\left ( 73.96 \right )^{2}+\left ( -22.66 \right )^{2}}$

= 77.35 m / s

The direction is Ф from X axis

$tan\phi =\frac{-22.66}{73.96}=-0.306$

Ф = -17° from + X axis or 343° from + X axis

7 0

4 years ago

To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 ✕ 10−10 m,

Elodia [21]

1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
$\Delta V =- \int\limits^{d}_{\infty} {E} \, dr$
Where E is the electric field generated by the proton, which is
$E=k_e \frac{q}{r^2}$
where $k_e=8.99\cdot10^9~Nm^2C^{-2}$ is the Coulomb constant and $q=1.6\cdot10^{-19}~C$ is the proton charge.
Replacing the electric field formula inside the integral, we obtain
$\Delta V =- \int\limits^{d}_{\infty} {k_e \frac{q}{r^2} } \, dr = k_e \frac{q}{d}= 27~V$

2) Then, we can calculate the work done by the electric field to move the electron (charge $q_e=-1.6\cdot10^{-19}C$ ) through this $\Delta V$ . The work is given by
$W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J$

5 0

3 years ago

Read 2 more answers

You are crouched at the "start" line. Bang! The race begins and 7 seconds later you end up 60 yards down the track. What was you

Nutka1998 [239]

Answer:

a = 7.35 ft / s²

Explanation:

For this exercise we must use the kinematics relations

x = v₀ t + ½ a t²

as the runner leaves the starting line his initial velocity is zero

x = ½ a t²