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dedylja [7]
2 years ago
7

Which component of the galaxy is shown in this image? Dust

Physics
1 answer:
Westkost [7]2 years ago
3 0

Answer:

stars

those are stars in the galaxy

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Rory uses a force of 25 N to lift her grocery bag while doing 50 J of work. How far did she lift the
Dennis_Churaev [7]

Answer:

W=Fd \\ 50 = 25d \\ d =  \frac{50}{25}  \\  \color{yellow} \boxed{d = 2m}

8 0
3 years ago
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
love history [14]

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

=0.001217 rad

The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

mλ = d sin θ ≈ dθ →

∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

6 0
3 years ago
Katherine johnson’s skill in what field led to nasa’s first moon voyage?
sertanlavr [38]
Answer: Katherine Johnson's knowledge of (mathematics) was instrumental in the return of the Apollo astronauts from the Moon to Earth.
4 0
2 years ago
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
Starting from rest, a basketball rolls from top of a hill to the bottom, reaching a translational speed of 6.8 m/s. Ignore frict
kkurt [141]

Answer:

Explanation:

for baseball

(a) Let the mass of the baseball is m.

radius of baseball is r.

Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy

T = 0.5 Iω² + 0.5 mv²

Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²

T = 0.83 mv²

According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.

m g h = 0.83 mv²

where, h be the height of the top of the hill.

9.8 x h = 0.83 x 6.8 x 6.8

h = 3.93 m

(b) Let the velocity of juice can is v'.

moment of inertia of the juice can = 1/2mr²

So, total kinetic energy

T = 0.5 x I x ω² + 0.5 mv²

T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²

m g h = 0.75 mv²

9.8 x 3.93 = 0.75 v²

v = 7.2 m/s

7 0
3 years ago
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