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3 years ago

Please answer faast!!!

Physics

1 answer:

Drupady [299]3 years ago

7 0

Answer:

Any medium or material of high refractive density e.g Water which refracts the light rays 1&2 away from their normal

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If u have a box with mass of 10kg and u push with a force of 200 N what is the acceleration of the box ?

AnnZ [28]

Answer:

20m/s²

Explanation:

force = mass*acceleration

acceleration = force ÷ mass

acceleration = 200N ÷ 10kg

acceleration = 20m/s²

5 0

3 years ago

List 5 differences between a compound and an element.

Anuta_ua [19.1K]

Answer:

1.An element consists of the same kind of atoms while a compound is composed of different kinds of atoms.

2.A monoatomic element cannot be split up physically or chemically but a compound can be split up into new substances by chemical means.

3.Only limited number of elements are known in an element whiles number of known compounds are quite large.

4.The property of an element is the property of the atom whiles the property of a compound is quite different from that of it's constituents.

5 . Elements are represented by symbols and numbers.for example Sodium is represented by Na whiles compounds are represented by their chemical formula.for example,Salt is represented by the formula NaCl.

5 0

3 years ago

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an

tensa zangetsu [6.8K]

The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.

5 0

3 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

3 years ago

4. Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 7.3 times tha

HACTEHA [7]

Answer:

Explanation:

Acceleration of particle A is 7.3 times the acceleration of particle B.

Let the acceleration of particle B is a, then the acceleration of particle A is

7.3 a.

Let the period of particle A is T and the period of particle B is 2.5 T.

Let the radius of particle A is RA and the radius of particle B is RB.

Use the formula for the centripetal force

$a=r\omega ^{2}=r\times \frac{4\pi^{2}}{T^{2}}$

So, $r = a\frac{T^{2}}{4\pi^{2}}$

The ratio of radius of A to the radius of B is given by

$\frac{R_{A}}{R_{B}}=\frac{a_{A}\times T_{A}^{2}}{a_{B}\times T_{B}^{2}}$

$\frac{R_{A}}{R_{B}}=\frac{7.3 a\times T^{2}}{a\times 6.25T^{2}}$

RA : RB = 1.17

3 0

3 years ago

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