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2 years ago

Plsss Help!!!!

Physics

2 answers:

jekas [21]2 years ago

5 0

I think the answer is d

Evgesh-ka [11]2 years ago

4 0

So the answer is c because what is happening is called “beta decay”. A, gamma doesn’t change the two numbers. B, alpha increases it with a hydrogen atom. and D would make the bottom number on the left of the equation 9.

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How much work is done when you push a crate horizontally with 150 N across a 13-m floor

Mila [183]

Work = (force) x (distance)

When a force of 150 N pushes through a distance of 13 meters,
it does

Work = (150 N) x (13 m) = 1,950 joules .

6 0

3 years ago

Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.

Yuki888 [10]

Scott traveled 10 miles

5 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

3 years ago

If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how

Arturiano [62]

Answer:

Vx= 11.0865(m/s)

Vy= 6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0

3 years ago