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ExtremeBDS [4]
3 years ago
7

Which of the following elements are most likely to have similar chemical properties?

Physics
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

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Is it true or not true
maria [59]

Answer:

true

Explanation:

6 0
3 years ago
A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the i
Marrrta [24]

Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

3 0
3 years ago
A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted
IrinaK [193]

Answer:

1.29649

488.08706 nm

6.14644\times 10^{14}\ Hz

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

\lambda_0 = 632.8 nm

From Snell's law we have the relation

n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649

Refractive index of the solution is 1.29649

Wavelength is given by

\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm

The wavelength of the solution is 488.08706 nm

Frequency is given by

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz

The frequency is 6.14644\times 10^{14}\ Hz

v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s

The speed in the solution is 231715700.28346 m/s

8 0
3 years ago
Total internal reflection is the complete reflection of a light ray _____.
Licemer1 [7]
B- Back into it's original medium.
Hope this helps!
7 0
3 years ago
Read 2 more answers
A large aquarium of height 5 m is filled with fresh water to a depth of D = 1.80 m. One wall of the aquarium consists of thick p
Damm [24]

To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be

(P-P_a)= \rho g x

This pressure acts on the strip of area

dA = 7.6dx

The force acting on that strip is given by,

dF = (P-P_a)dA

dF = \rho g xdA

dF = 7.6 \rho g x dx

To evaluate the force, we will then consider the integral of the pressure as a function of the Area, or the integral of the previously found terms.

F = \int_0^x 7.6 \rho g x dx

F = 3.8 \rho g x^2

Evaluating at the initial depth of 1.8m and the final depth of 4.4 we have then that,

F_{1.8} = 3.8 (998)(9.8)(1.8)^2 = 120416.28N

F_{4.4} = 3.8(998)(9.8)(4.4)^2 = 719524.46N

Therefore the Net force will be

F = F_{4.4}-F_{1.8}

F = 719524.46-120416.28

F = 599108.18N

8 0
4 years ago
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