All categories

3 years ago

What is the length x of the side of the triangle below

Physics

2 answers:

Shalnov [3]3 years ago

8 0

Answer:

you didn't post any triangles. Thus, the question could not be answered.

alexdok [17]3 years ago

7 0

Answer:

The given triangle is a right triangle.

The length of the hypotenuse is 2.52.5 units.

The given angle is 39\degree39° .

The unknown side xx is adjacent to the given angle.

We can therefore use the cosine function to find xx .

Recall the mnemonics CAH.

This means that;

\cos\theta=\frac{Adjacent}{Hypotenuse}cosθ=

Hypotenuse

Adjacent

This implies that,

\cos39\degree=\frac{x}{2.5}cos39°=

2.5

We multiply both sides by 2.52.5 to get,

2.5\cos39\degree=x2.5cos39°=x

x=2.5(0.7771)x=2.5(0.7771)

We evaluate to get

x=1.94x=1.94

Therefore the length xx of the side of the triangle is 1.91.9 units to the nearest tenth.

You might be interested in

Sug . To repeat the experiment what is it?

statuscvo [17]

Answer:

....more foam

Explanation:

7 0

3 years ago

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

2 years ago

What is the idea behind the law of multiple proportions?

Akimi4 [234]

Answer:

The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed

Explanation:

5 0

3 years ago

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

Required:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

Solution:

Angular velocity (Δω):

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

Average angular acceleration (αav):

αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0

2 years ago

Read 2 more answers

As you tune your radio to a higher frequency, from 88.5 FM to 103.3 FM, the wavelength of the wave that you are tuning in

olga_2 [115]

the relation that relates the speed of wave, frequency of wave and wavelength is given as

wavelength = $\frac{speed}{frequency}$