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3 years ago

Helppppppppppppppppppppppppppppppppppppppppppppppp

Physics

1 answer:

marta [7]3 years ago

8 0

Answer:

My logic:

Heliocentrism is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Universe.

Hope this helps!

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Please helppp....What are 5 examples of real science

valentina_108 [34]

Answer:

dynamics

kinematics

fluid mechanisms

earth science and

physics

8 0

3 years ago

a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box

tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , $\theta=39^o$ .

Coefficient of kinetic friction , $\mu=0.16$ .

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

$W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.$

Hence , this is the required solution.

6 0

3 years ago

A 5.17 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The othe

Andreyy89

Answer:

v=0.816 m/s

Explanation:

The force of the spring and the motion of the block are in equilibrium so without any force of friction the motion is

$E_{s}=E_{k}$

$\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2$

First determinate the constant of the spring that produce the kinetic energy of the bloc

$k=\frac{m*v^2}{d_{s}^2}$

$k=\frac{5.17kg*(1.30\frac{m}{s})^2}{0.102^2m}$

$k=839.8 \frac{kg}{s^2}$

Now the motion with the force of friction in the kinetic

$E_{s}=E_{k}-W_{k}$

$\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2-u*m*g$

Resolve to v

$v^2=\frac{(k*d_{s}^2)+(2*g*u)}{m}$

$v=\sqrt{\frac{839.8*(0.102m)^2-2*9.8*0.270}{5.17}}$

$v=0.81 \frac{m}{s}$

3 0

3 years ago

The energy levels of one‑electron ions are given by the equation En=(−2.18×10−18 J)(Z^2/n^2) where Z is atomic number and n is t

Alexxandr [17]

Answer:

$\lambda=4.86*10^{-7}m$

Explanation:

Using the given equation, we calculate the energy associated with the excited state $n_i=8$ and $n_f=4$

$E_n=\frac{-2.18*10^-18J(Z^2)}{n^2}$

Helium has an atomic number (Z) equal to 2, for n=8:

$E_8=\frac{-2.18*10^{-18}J(2^2)}{8^2}\\E_8=-1.36*10{-19}J$

For n=4:

$E_4=\frac{-2.18*10^{-18}J(2^2)}{4^2}\\E_4=-5.45*10{-19}J$

When an electron jumps from an energy level with greater energy $E_i$ to one with lower energy $E_f$ the wavelength of the emitted photon is given by:

$\lambda=\frac{hc}{E_i-E_f}$

h is the Planck constant and c the speed of light in vaccum. So, we have:

$\lambda=\frac{hc}{E_8-E_4}\\\lambda=\frac{6.63*10^{-34}J \cdot s(3*10^8\frac{m}{s})}{-1.36*10{-19}J-(-5.45*10{-19}J)}\\\lambda=4.86*10^{-7}m$

5 0

3 years ago

A machining company needs to manufacture more than 20 fixtures in a day. The company uses five identical machines to make the fi

balu736 [363]

a put a my boy

Explanation:

5x5=25 so it would be a

5 0

2 years ago