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2 years ago

10

Solve ~

13x - 26 + 39 = 0" align="absmiddle" class="latex-formula">

ty! :)

2 answers:

Natalka [10]2 years ago

4 0

Explanation:

13x - 26 + 39 = 0

13x=26-39

13x=-13

x= -1

liq [111]2 years ago

3 0

$13x-26+39=0\\\\\implies 13x +13 = 0\\\\\implies 13x = -13\\\\\implies x = -\dfrac{13}{13}\\\\\implies x = -1$

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What is an electric field? How<br> do you calculate the electric<br> field due to a point charge?

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An electric field is an electric property associated with each point in space when charge is present in any form.

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2 years ago

What is the frequency of radiation whose wavelength is 0.80 nm ? express the frequency in inverse seconds to three significant d

Mashcka [7]

The relationship between the frequency (f) of an electromagnetic wave, the speed of the wave (which is the speed of light, c) and its wavelength $\lambda$ is given by
$f= \frac{c}{\lambda}$

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3 years ago

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

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$\dfrac{1}{2}mv^2=mgl$

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Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

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The initial speed of the block $u_{2}=0$

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Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

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8 0

3 years ago

The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

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3 years ago

Read 2 more answers

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$

Therefore, the angle of projection is 12.26⁰.

6 0

2 years ago