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Sunny_sXe [5.5K]
3 years ago
6

Solve 3* +5-220t = 0​

Physics
1 answer:
slega [8]3 years ago
4 0

Answer:

t = 27.5

Explanation:

3 + 5 -220t = 0

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

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You place an object 30.0 cm from a concave mirror of 15.0 cm focal length. The object is 1.8 cm tall. Find the image position an
Softa [21]

Answer:30 cm

Explanation:

Given

object distance u=30\ cm

focal length of concave mirror f=15\ cm

height of object h_o=1.8\ cm

Using mirror formula

\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}

\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{30}

\Rightarrow \frac{1}{v}=\frac{2-1}{30}

\Rightarrow v=30\ cm

and magnification is

\Rightarrow m=\frac{-v}{u}=\frac{h_i}{h_o}

\Rightarrow \frac{-30}{30}=\frac{h_i}{1.8}

\Rightarrow h_i=-1.8\ cm

So height of object is same as of object .

Position :image is formed at the spot where object is placed.

4 0
3 years ago
Un cuerpo de 60 kg se encuentra a una distancia de 3.5 m del otro cuerpo, de manera que entre ellos se produce una fuerza de 6.5
aev [14]

Answer:

1989.6Kg

Explanation:

The computation of the mass of the other body is given below:

As we know that

F = G × m1 × m2 ÷ r²

Here the G would have the constant value i.e. 6.67 × 10^-11Nm² / kg².  

Now

6.5 × 10^-7N = 6.67 × 10^-11Nm² / kg² × 60Kg × m2 / (3.5m) ²

m2 = (F × r²) / (G × m1)

m2 = (6.5 × 10^-7N × (3.5m) ²) ÷ (6.67 × 10^-11Nm² / kg² × 60Kg)

= 1989.6Kg

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3 years ago
The impulse experienced by a body is equivalent to the body’s change in?
vovangra [49]
<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

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5 0
4 years ago
The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 9.00×1010 atoms of zinc e
KengaRu [80]

Answer:

8.4\cdot 10^{-8}J

Explanation:

The energy emitted by a single photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

For the photons emitted by the zinc atoms,

\lambda = 214 nm = 214 \cdot 10^{-9} m

So the energy of a single photon emitted is

E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{214\cdot 10^{-9}}=9.3\cdot 10^{-19}J

And since the number of atoms is

N=9.0\cdot 10^{10}

The total energy emitted will be

E=NE_1 = (9.0\cdot 10^{10})(9.3\cdot 10^{-19})=8.4\cdot 10^{-8}J

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4 years ago
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