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4 years ago

Scientific notation for 836

Physics

2 answers:

JulijaS [17]4 years ago

8 0

Answer:

8.36e2

Explanation:

Use a scientific calculator

tensa zangetsu [6.8K]4 years ago

6 0

Its notation is written as :

8.36 x 10^2

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Why isn't the wave nature of light (visible electromagnetic radiation) obvious to us?

vichka [17]

<span>The wavelength of light measures in at around 20 millionths of an inch, making it incredibly small, too small to be seen with the naked eye an this is why it is not obvious to us. This is the type of thing that must be seen under an extremely powerful tool.</span>

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3 years ago

If it takes 5 seconds for the projectile to reach its highest point, what is the total

Ganezh [65]

10 seconds! There you go

5 0

3 years ago

A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1

const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0

4 years ago

A meteorologist measures the angle of elevation of a weather balloon as 41 degrees. A radio signal from the balloon indicates th

Katena32 [7]

This problem we can represent as right triangle where we can use basic trygonometric functions.

in this case we have largest side (hypothenuse) of the triangle and we have one angle. The angle we got is the one between side of triangle that is horizontal with the ground and the hypothenuse, meaning that we need sin function to connect relation between angle, known side lenght and unknown side lenght.

sin(alpha) = (side of triangle oposite from observed angle)/(hypothenuse)

sin(41) = x/1503

x = 1503*sin41 = 986.06 meters

Answer is 986.06 meters

8 0

3 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago