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galben [10]
3 years ago
5

What is the value of x in the equation below 1+2e^x+1=9

Physics
1 answer:
GuDViN [60]3 years ago
3 0
<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of x:

1+2e^x+1=9   (1)

Firstly, we have to clear x:

2e^x=9-1-1  

2e^x=7  

e^x=\frac{7}{2}     (2)

Applying<u> Natural Logarithm</u> on both sides of the equation (2):

ln(e^x)=ln(\frac{7}{2})     (3)

xln(e)=ln(\frac{7}{2})     (4)

According to the Natural Logarithm rules xln(e)=x, so (4) can be written as:

x=ln(\frac{7}{2})     (5)

Finally:

x=1.252    

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lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0
3 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
Equation for pressure at a depth H inside a fluid PLSS URGENTT
Blababa [14]

Answer:

We begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

4 0
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What time period is the child with two angels from?
oee [108]
I believe is was 1465. hope this helped!  
5 0
3 years ago
Consider a satellite in a circular orbit around the Earth. If it were at an altitude equal to twice the radius of the Earth, 2RE
Elenna [48]

Answer:

v=\sqrt{\frac{gR_E}{2}}

Explanation:

Satellites experiment a force given by Newton's Gravitation Law:

F=\frac{GMm}{r^2}

where M is Earth's mass, m the satellite's mass, r the distance between their gravitational centers and G the gravitational constant.

We also know from Newton's 2nd Law that <em>F=ma, </em>so putting both together we will have:

ma=\frac{GMm}{r^2}

a=\frac{GM}{r^2}

If we are on the surface of the Earth, the acceleration would be g and r=R_E (Earth's radius):

g=\frac{GM}{R_E^2}

Which we will write as:

gR_E^2=GM

If we are on orbit the acceleration is centripetal (a=\frac{v^2}{r}), so we have:

\frac{v^2}{r}=a=\frac{GM}{r^2}=\frac{gR_E^2}{r^2}

v^2=\frac{gR_E^2}{r}

v=\sqrt{\frac{gR_E^2}{r}}

And if this orbit has a radius r=2R_E we have:

v=\sqrt{\frac{gR_E^2}{2R_E}}=\sqrt{\frac{gR_E}{2}}

3 0
3 years ago
Read 2 more answers
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