1 question, brainliest for whoever gets it right.

natita [175]

C is probably the correct one

3 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$ kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$ = $\frac{100}{3}$ m/s

and $V_y=\frac{160}{3}$ m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

= $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

= 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$ = 1.6

therefore, $\theta = \tan^{-1}1.6$

= $58 ^{\circ}$ from x-axis

4 0

3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago

An airplane weighing 11,000 N climbs to a

Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

$W=(mg)\Delta h$

where

mg = 11,000 N is the weight of the airplane

$\Delta h = 1.6 km = 1600 m$ is the change in height

Substituting,

$W=(11,000)(1600)=17.6\cdot 10^6 J$

Now we can calculate the power delivered, which is given by

$P=\frac{W}{t}$

where

$W=17.6\cdot 10^6 J$ is the work done

$t=9.8 min \cdot 60 = 588 s$ is the time taken

Substituting,

$P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W$

Finally, we can convert the power into horsepower (hp), keeping in mind that

$1 hp = 746 W$

Therefore,

$P=\frac{2.99\cdot 10^4}{746}=40.1 hp$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago

Harmfful effect of earthquake

Snowcat [4.5K]

Answer:

Death, Destruction, Loss of home

5 0

3 years ago

Read 2 more answers