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3 years ago

Rays of light strike a bumpy road and are reflected.

Physics

2 answers:

Masteriza [31]3 years ago

8 0

Answer:

D. Each will reflect at an angle equal to its angle of incidence.

Explanation:

ser-zykov [4K]3 years ago

6 0

Answer:

last option is the correct one

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Which instrument is used to compare the mass of an object

Vladimir [108]

Answer:

A) a balance

Explanation:

The balance compares an object with a known mass to the object in question. Different types of balances include digital scientific balances and beam balances, such as a triple beam balance.

5 0

3 years ago

A large aquarium of height 6 m is filled with fresh water to a depth of D = 1.50 m. One wall of the aquarium consists of thick p

Vesnalui [34]

Answer

given,

height of aquarium = 6 m

Depth of fresh water = D = 1.50 m

horizontal length of the aquarium(w) = 8.40 m

total force increased when liquid is filled to depth = 4.30 m

g = 9.81 m/s²

ρ = 998 Kg/m³

force in the aquarium.

dF = PdA

$F = \int PdA$

$F = \int \rho\ g\ y\ (wdy)$

$F = \rho\ g\ w \int ydy$

$F = \rho\ g\ w\dfrac{y^2}{2}$

$F = \dfrac{\rho\ g\ w\ y^2}{2}$

At D = 1.5 m

$F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}$

F = 9.08 x 10⁴ N

At D = 4.30 m

$F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}$

F = 7.46 x 10⁵ N

Total force on the wall increased by

ΔF = 74.6 x 10⁴ - 9.08 x 10⁴

ΔF = 65.52 x 10⁴ N

5 0

3 years ago

using a weston cadmium cell of 1.0283 v and a standard resistance of .1ohm a potentiometer was adjusted so that 1.0183 m was equ

anyanavicka [17]

Answer:

Explanation:

the value is -8 cm

3 0

2 years ago

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

3 years ago

point A is at the bottom of a rough plane which is inclined at angle tita to the horizontal . A body of mass M is projected from

ziro4ka [17]

<h2>The work done , when body moves along the plane </h2>

Explanation:

A body is projected from bottom of the inclined plane . When body is going up the plane .

The downward force = m g sinθ is developed due to its weight

As body is moving upwards , the force of friction will act downwards

The force of friction = μ R

here μ is the coefficient of friction ans R is the normal reaction

Thus force of friction f = μ mg cosθ

Let the acceleration upwards is a

The upward force required = m a

Thus m a = mg sinθ + μ mg cosθ

or acceleration a = g ( sinθ + μ cosθ )

The work done in moving upwards W = F S

Thus W = mg ( sinθ + μ cosθ ) S

here S is the displacement on the plane

When body moves down , the force of friction acts upwards

Thus m a = m g ( sinθ - μ cosθ )

The work done W = m g ( sinθ - μ cosθ ) S

As the body is projected with velocity u

which can be calculated by the relation v² - u² = - 2 a X

Here v = 0 at the highest point

Thus u = $\sqrt{2ax}$

here a = g ( sinθ + μ cosθ )

Similarly , when it moves down , the initial velocity u = 0

Thus v² - 0 = 2 a x

or v = $\sqrt{2ax}$

here a = g ( sinθ - μ cosθ )

3 0

3 years ago