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3 years ago

A rocket is launched with a velocity of 30 m/s at 40° to the horizontal. How far will it

Physics

1 answer:

steposvetlana [31]3 years ago

8 0

Answer:

A. Rocket A will travel farther horizontally than rocket B.

Explanation:

This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.

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A student is skateboarding down a ramp that is 6.0 m long and inclined at 180 with respect to the horizontal. The initial speed

dlinn [17]

Answer:

The speed at the bottom of the ramp is 2.6m/s

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

Notice that it is necessary to found the acceleration that can be done by means of Newton's second law:

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

<em>Force in the x axis:</em>

$F_{x} = W_{x}$

The component of the weight in the x axis can be gotten by means of trigonometric:

$\frac{OC}{H} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$F_{x} = mgsen \theta$ (3)

<em>Forces in the y axis: </em>

$F_{y} = N - W_{y}$ (4)

The component of the weight in the y axis can be gotten by means of trigonometric:

$\frac{AC}{H} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

$N = mgcos \theta$

Therefore, equation 4 can be rewritten as:

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$ (5)

Then, replacing equation 3 and equation 5 in equation 2 it is gotten:

$mgsen \theta + 0 = ma$

$mgsen \theta = ma$ (6)

However, a can be isolated from equation 6

$a = \frac{mgsen \theta}{m}$

$a = gsen \theta$ (7)

Finally, equation 7 can be replaced in equation 2:

$v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}$

$v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}$

$v_{f} = 2.6m/s$

Hence, the speed at the bottom of the ramp is 2.6m/s

7 0

3 years ago

Cuál es la diferencia entre fricción beneficiosa y perjudicial? Pon 5 ejemplos de cada uno.

Rus_ich [418]

Answer:

Difference is given below.

Explanation:

The main difference between beneficial and harmful friction is that beneficial friction is necessary for performing different activities while on the other hand, harmful effect of friction is destroy different parts of products and machines. examples of beneficial friction are walking, holding things, rubbing hands to produce heat, running etc whereas examples of harmful friction are destruction of sole, slipping, tearing of machine's part, Wet roads and Mudslides etc.

7 0

3 years ago

Small birds can migrate over long distances without feeding, storing energy mostly as fat rather than carbohydrate. (Figure 1) F

Katen [24]

Answer: 3.1107grams of fat.

Explanation:The speed to cover 800 km in 20 hours will be 40km/hr, Because

Speed =Distance/Time.

This is equal to 72,000 seconds. At a power consumption of 1.7 W, the bird will consume (1.7 X 72,000) =122,400 J of energy.

A gram of fat contains about (9.4 Calories X 4186 J/Calorie) ,=39348 joules of energy, the bird will need (122,400 / 39,348) = 3.1107 grams of fat.

4 0

3 years ago

Which part of the electromagnetic spectrum has the longest wave lengths

Lana71 [14]

Radio Waves :)) i’m pretty confident in that

8 0

3 years ago

The star Betelgeuse is 6.1 x 10^18 m away from Earth. How old is the light we see from that star when it reaches us? There are 3

castortr0y [4]

Answer:

635 years old

Explanation:

The light reaching the earth from the sun will travel at a speed called the speed of light, and this has a universal value of 3 × 10⁸ m/s. Bearing this in mind, let us calculate the age of the light reaching the Earth from the sun:

Distance of star from Earth = 6.1 × 10⁸m

Speed of light = 3 × 10⁸ m/s

We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

Distance = speed × time

6.1 × 10⁸ = 3 × 10⁸ × time

$time = \frac{6.1 \times 10^{18}}{3 \times 10^8}$

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

$\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds$

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

∴ The number of years travelled by the light from the star:

$3.2\ \times 10^7\ seconds = 1\ year\\1\ second = \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}$

please note that:

2.03 × 10¹⁰ = 20300000000

3.2 × 10⁷ = 32000000

$\therefore \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}\\= \frac{20300000000}{32000000} \\\\= \frac{20300}{32} \\= 634.347\ years\\$

The closest answer in the option is 635 years, and we are short of this by some points due probably to approximations in the calculation.

8 0

3 years ago