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3 years ago

Two angles are supplementary. The first angle measures 60° what is the measurement of the second angle?

1 answer:

3 0

<span> if 2 angles are supplementary, then their sum is 180 degrees
so have x be the measure of the second angle and the measure of the first one
then,
x+y=180
x+60=180
x=180-60=120
soi think it would be c</span>

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A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 Newtons over Coulombs.. Determin

shtirl [24]

The magnitude of the electric force on the charge is 5 N.

<h3>Magnitude of force on the charge</h3>

The magnitude of force on the charge is calculated as follows;

F = Eq

where;

E is electric field
q is magnitude of the charge

F = 100 N/C x 0.05 C

F = 5 N

Thus, the magnitude of the electric force on the charge is 5 N.

Learn more about electric force here: brainly.com/question/20880591

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1 year ago

You walk into the kitchen and see a broken egg on the floor. Which of the following is an inference you can make based on this o

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3 years ago

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Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

<u>Note that the acceleration is negative because is a braking process.</u>

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3 years ago

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A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo

user100 [1]

Answer:

$\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

Explanation:

The total force on the particle is given by

$\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}$

Then, by replacing we have:

$q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$