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Strike441 [17]
3 years ago
6

(HELP PLEASEE) An object is 40.0 cm from a

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

-18

Explanation:

acellus use the formula of magnification and watch your minus sign

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a horse runs with an initial velocity of 11m/s and slows to 5.2 m/s over a time interval of 3.1 s what is the horse's average ac
Katarina [22]

Answer:

a = change in v / change in time

= (5.2 - 11) / 3.1

= -1.87 m/s^2

Explanation:

6 0
3 years ago
How do you solve <img src="https://tex.z-dn.net/?f=4x%5E%7B3%7D" id="TexFormula1" title="4x^{3}" alt="4x^{3}" align="absmiddle"
Vlad1618 [11]

Hello There!

Here's a explanation!

Let's solve your equation step-by-step.

4x^3=2x^-^1

4x^3=\frac{2}{x}

Step 1: Multiply both sides by x.

4x^4=2

\frac{4x^4}{4} =\frac{2}{4}

(Divide both sides by 4).

x^4=\frac{1}{2}

x=+(\frac{1}{2} )^(^\frac{1}{4} ^)

Take the root.

ANSWER!

x=0.840896 Or x=-0.840896

Hopefully, this helps you!!

AnimeVines

8 0
3 years ago
It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
Maru [420]

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

 \phi=60(150)^2cos180+100(150)^2*cos0

 \phi=-9*10^5

Generally Quantity of charge  q is mathematically given as

 q=\varepsilon _0 *\phi

 q=8.85*10^-^1^2 *9*10^5

 q=7.965*10^-^6C

5 0
2 years ago
What is the unit for magnitudes in astronomy?
Sonja [21]
The unit is light years or Ly
3 0
3 years ago
An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a In turn it forms an α-angle α
xxMikexx [17]
Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta -  \alpha ) \frac{d}{u} - h = 0
4.9[ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha }  ]^{2} - v_{0} sin(\theta -  \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha } ] - X sin \alpha  = 0
Hence obtain
aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta -  \alpha )}]^{2} \\  b = cos \alpha \,  tan(\theta -  \alpha ) + sin \alpha
The non-triial solution for X is
X= \frac{b}{a}

Answer:
X= \frac{sin \alpha  + cos \alpha  \, tan(\theta -  \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta -  \alpha )}  ]^{2}}

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

Answer:  X = 101.5 m

7 0
3 years ago
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