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allsm [11]
3 years ago
11

a moving electron accelerates at 5200m/s^2 in a 55.0 direction. After 0.350as,it has a velocity of 6598m/s in a -20.5 direction.

What is the y-component of the initial velocity?
Physics
1 answer:
Lady bird [3.3K]3 years ago
6 0

Answer:

  -3802 m/s

Explanation:

The y-component of the final velocity is ...

  (6598 m/s)·sin(-20.5°) ≈ -2310.7 m/s

The y-component of the velocity due to acceleration is ...

  (5200 m/s²)(0.350 s)sin(55°) ≈ 1490.9 m/s

Then the initial velocity in the y-direction is found from ...

  initial velocity + change in velocity = final velocity

  initial velocity = (final velocity) - (change in velocity)

  = -2310.7 m/s - 1490.9 m/s ≈ -3802 m/s

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The electrons of carbon and hydrogen are _____________ in this molecule of methane.
nlexa [21]

Answer:

The answer is C

7 0
3 years ago
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
Indiana jones (83.5 kg) is running 3.75 m/s when he jumps in a stationary 312 kg mine cart. what is their joint velocity afterwa
Lubov Fominskaja [6]

Answer:

.7917 m/s

Explanation:

This is a conservation of momentum question. You have an object initially at rest (cart) so that object is initially at 0 momentum. Indiana Jones is 83.5 kg and running 3.75 m/s so he starts with a momentum of 313.125 kg * m/s because momentum is equal to mass * velocity. Once the person jumps in the cart, the cart and the person can be considered one object and by conservation of momentum, the momentum of the Indiana-cart system is equal to 313.125 kg * m/s. By that, we can set that momentum equal to the combined mass * joint velocity. So 313.125 = (83.5kg + 312kg) * joint velocity. Then just solve for the velocity. The answer should be smaller than the intial velocity of the person of 3.75 m/s because the mine cart is HUGE at 312kg.

3 0
3 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

8 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
3 years ago
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