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3 years ago

Please help with 3 questions about acceleration

Physics

1 answer:

Goryan [66]3 years ago

5 0

A great, helpful, useful definition of acceleration is

<em>A = (change in speed) / (time for the change)</em> . <== you should memorize this

This simple tool will directly solve all 3 problems.

The REASON for assigning these problems for homework is NOT to find the answers. It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do. Noticed that if you get the answers from somebody else, you lose all of these benefits.

The only wrinkle anywhere here is in #3, because when you use this definition, the unit of time has to be the same in both the numerator and the denominator.

So for #3, you have to EITHER change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.

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A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

Lorico [155]

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

3 0

3 years ago

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

3 years ago

If we have less power, we most likely have

boyakko [2]

The same amount of work being done over a long period of time!

6 0

3 years ago

Determine if the data are qualitative or quantitative.

Amanda [17]

Answer:

Qualitative, Quantitative, Qualitative, Quantitative, and Qualitative.

Explanation:

7 0

2 years ago

Read 2 more answers

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago