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VMariaS [17]
2 years ago
12

What is the name of the ionic compound Cs2S

Physics
2 answers:
tekilochka [14]2 years ago
4 0

Answer:cesium sulfide

Explanation:

Ugo [173]2 years ago
3 0
The answer is cesium oxide
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child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
2 years ago
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail
Alik [6]

Answer:

1.34\cdot 10^{-16} C

Explanation:

The strength of the electric field produced by a charge Q is given by

E=k\frac{Q}{r^2}

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C

4 0
2 years ago
If the work required to move a 0.35 c charge from point a to point b is 125 j, what is the potential difference between the two
valentina_108 [34]
Using W = qV  =>  V = W/q


7 0
2 years ago
PLEASE HELP ME WITH A PHYSICS QUESTION!!!!!!!!!!
stira [4]

Answer:

C. 3.00 s

Explanation:

Given:

Δy = 1.80 m − 46.0 m = -44.2 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 3.00 s

6 0
3 years ago
Read 2 more answers
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
2 years ago
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