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3 years ago

2 QUESTIONS HELP PLZ...:/

Physics

1 answer:

algol [13]3 years ago

5 0

The first one is: head
Second one is: 10 trillion km

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The zone along the southern margins of the Sahara is called the

Jobisdone [24]

We have to choose the correct name for the zone along the southern margins of the Sahara. Laterite is a soil in hot and wet tropical areas. Savanna is the tropical grassland. It has tropical savanna climate. Veldt is name for the areas in the South Africa. Finally, the Sahel is the zone along the south margins of the Sahara. It has a semi-arid climate. The Arabic word "sahel" means "coast". Answer: C. Sahel. <span> </span>

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3 years ago

Which planet moves the fastest and which planet moves slowest? a, b, c, d, e, or f. Explain

Alisiya [41]

D is the point where the planet moves the fastest. This is because it is in the perihelion, where the planet is moving at it’s fastest pace

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1 year ago

shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell

earnstyle [38]

Divide 14 by 6 and there is your answer with the unit of m

8 0

3 years ago

Read 2 more answers

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

o-na [289]

The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.

(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m

Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.

(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ

θ=37.69 degrees

Hopes this helps!

P.S. I hope this is right. If not sorry in advance.

3 0

3 years ago