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3 years ago

What is the value of x in the triangle? a 45-45-90 triangle with leg length x and hypotenuse length 4

Physics

2 answers:

mrs_skeptik [129]3 years ago

5 0

Answer:

$x = 2\sqrt 2$

Explanation:

Given

$Hypotenuse = 4$

Required

Find x

Since the triangle is a 45-45-90 triangle, the following relationship exists

$x^2 + x^2 = 4^2$ --- i.e. the other legs are equal

So, we have:

$2x^2 = 16$

Divide both sides by 2

$x^2=8$

Take square roots of both sides

$x = \sqrt 8$

Simplify

$x = 2\sqrt 2$

myrzilka [38]3 years ago

4 0

Answer:

7√2

Explanation:

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a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick

Svetradugi [14.3K]

The formula for potential energy is
E(p) = mgh

(Mass x gravity x height)

Therefore energy = (5.3)(9.8)(6.6)
= 342.8 J

How did I get 9.8?
9.8 is the constant for gravity

8 0

4 years ago

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago

How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v

777dan777 [17]

The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
$W= qV_i - qV_f$
where q is the charge of the proton, $q=1 e = 1.6\cdot 10^{-19}C$ , with $e$ being the elementary charge, and $V_i = +125 V$ and $V_f = -55 V$ are the initial and final voltage.

Substituting, we get (in electronvolts):
$W=e(125 V-(-55 V))=180 eV$
and in Joule:
$W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J$

5 0

3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

4 years ago

Help me find the acceleration

ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the timeless suvat equation is likely what you need for this question.

In the timeless suvat equation,

$a = \dfrac{v^2 - u^2}{2\; x}$

where

$a$ is the acceleration of the car;
$v$ is the final velocity of the car;
$u$ is the initial velocity of the car; and
$x$ is the displacement of the car.

Note that v and u are velocities. Make sure that you include their signs in the calculation.

In this question,

$a$ is the unknown;
$v = -10.9 \; \text{m} \cdot \text{s}^{-2}$ ;
$u = -27.7 \; \text{m} \cdot \text{s}^{-2}$ ; and
$x = - 105 \; \text{m}$ .

Apply the timeless suvat equation:

$a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}$ .

The value of $a$ is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0

3 years ago