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3 years ago

Two laws are described below:

Physics

2 answers:

taurus [48]3 years ago

8 0

Answer:

i beleive the answer is B Both Law A and Law B are scientific laws.

Explanation:

liubo4ka [24]3 years ago

3 0

Answer:

Explanation:

I can conclude that this means that the law can be broken under certain coditions as long as its not focused on a natural phenomonon

and a pheononmono is a fact or situation that is observed to exist or happen, especially one whose cause or explanation is in question.

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Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

3 years ago

A 15kg ball accelerates at a rate of 3m/s/s. What force was required?

hoa [83]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 3

We have the final answer as

Hope this helps you

3 0

3 years ago

A sled of mass 15 kg slides across the ground with a friction force of 80.85 N. What is k between the sled and the ground?

Dvinal [7]

The expression for the frictional force between the sled and the ground is:
$F=k m g$
where $k$ is the coefficient of friction, m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration.

The friction force in our problem is F=80.85 N. The mass of the object is m=15 kg. Re-arranging the formula, we can find the value of k:
$k= \frac{F}{mg}= \frac{80.85 N}{(15 kg)(9.81 m/s^2)}=0.55$

8 0

3 years ago

Read 2 more answers

How can you define a solution to an equation?

sleet_krkn [62]

A solution is a value or a collection of values.. when substituted for the unknowns, the equation become an equality.
Example : x + 2 = 7
When we out the 5 in place of x we get: 5 + 2 = 2

8 0

3 years ago