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2 years ago

If

Physics

2 answers:

olga2289 [7]2 years ago

6 0

Answer:

I don't understand that

bezimeni [28]2 years ago

3 0

Answer:

I don't understand

Explanation:

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A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

2 years ago

When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas

Andrew [12]

Answer:

The source of cosmic background radiation filled the entire universe.

Explanation:

D:The source of cosmic background radiation filled the entire universe.

7 0

2 years ago

Read 2 more answers

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

3 years ago

Explain the term "avalanche state"

vagabundo [1.1K]

Answer:

The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.

Explanation:

5 0

3 years ago

At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−

vredina [299]

Answer:

a) Δx = 49.23 mi , b) Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

a = dv / dt

∫dv = ∫ a dt

we integrate and evaluators

v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

1+ 8t = u

8 dt = du

v - v₀ = -1280 ∫ u⁻³ du / 8

v -v₀ = -1280 / 8 (-u⁻²/2)

v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

v- 80 = 80 [(1 + 8t)⁻² - 1]

v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

v = dx / dt

dx = vdt

x-x₀ = 80 ∫ (1-8t)⁻² dt

x-x₀ = 80 ∫ u⁻² dt / 8

x-x₀ = 80 (-1 / u)

x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

x = x₀

t = 0.2 h

x-x₀ = -80 [1 / (1 +8 02) -1]

x-x₀ = 49.23

displacement is

Δx = x (0.2) - x (0)

Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

x- x₀ = -80 [1 / (1+ 8 0.4) -1]

x-x₀ = 55 mi

Δx = x (0.4) - x (0.2)

Δx = 55 - 49.23

Δx = 5.77 mi

3 0

3 years ago