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3 years ago

7

a wheelchair ramp for a business cannot be steeper than 5° a similar ramp for a home can be 10° what is the difference in degree

s of these two ramps? Explain.

1 answer:

Marina86 [1]3 years ago

4 0

It’s a 5° difference because 10-5 = 5

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The radioactivity of an atom is dependent on __________

Lana71 [14]

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</span><span>
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8 0

3 years ago

How can you increase the rate of a reaction? Select all that apply. *

sergij07 [2.7K]

Answer:

- Increase the concentration of the reactants

-increase the temperature

- introduce a catalyst

6 0

3 years ago

A theme park creates a new kind of water wave pool with large waves caused by constructive interference. There are two wave gene

melomori [17]

Answer:

10.0 m

Explanation:

Since there is no amplitude at the point of the swimmer, we have destructive interference.

So, the path difference ΔL = L₂ - L₁ where L₁ = swimmer's shorter distance from one generator = 9.0 m and L₂ = swimmer's longer distance from the other generator = 14.0 m. ΔL = 14.0 m - 9.0 m = 5.0 m

Also, since we have destructive interference, ΔL = (n + 1/2)λ where n = number of wavelengths and λ = wavelength of waves

For maximum wavelength, n = 0

So, ΔL = (n + 1/2)λ

ΔL = (0 + 1/2)λ

ΔL = λ/2

λ/2 = ΔL

λ = 2ΔL

λ = 2 × 5.0 m

λ = 10.0 m

So, the longest wavelength that will produce this interference pattern is λ = 10.0 m

8 0

3 years ago

Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0

4 years ago

Three objects lie in the x, y plane. Each rotates about the z axis with an angular speed of 5.58 rad/s. The mass m of each objec

QveST [7]

Answer:

a) V1=11.05m/s V2=92.07m/s V3=17.24m/s

b) KE = 16238.26J

Explanation:

For tangential speeds:

$V1 = \omega*R1=5.58*1.98=11.05m/s$

$V2 = \omega*R2=5.58*16.5=92.07m/s$

$V3 = \omega*R3=5.58*3.09=17.24m/s$

For the kinetic energy, it can be calculated as:

$KE=1/2*\omega^2*(I1+I2+I3)$

Where:

$I1 = m1*R1^2=5.46*1.98^2=21.4kg.m^2$

$I2 = m2*R2^2=3.64*16.5^2=990.99kg.m^2$

$I3 = m3*R3^2=3.21*3.09^2=30.65kg.m^2$

So,

$KE=1/2*5.58^2*(21.4+990.99+30.65)$

KE=16238.26J

4 0

3 years ago