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vivado [14]
3 years ago
5

player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the hor

izontal velocity component of the ball to the nearest tenth of a m/s?
Physics
2 answers:
seropon [69]3 years ago
7 0

Answer

22.5 m/s

Explanation

We shall use the trigonometric ratio cosine to find the horizontal component.

cos = adjacent/hypotenuse

Adjacent is the horizontal and hypotenuse is the fly speed.

cos 30° = horizontal / 26

horizontal velocity = 26 × cos 30°

                                = 26 × 0.866

                                = 22.5166

                                 = 22.5 m/s

Ad libitum [116K]3 years ago
4 0

Solution

In this Question we have given,

Velocity, V=26\frac{m}{s}

angle,α=30 degrees

We have to find horizontal component of velocity,

V_{x}=V\times Cos\alpha

V_{x}=26\times Cos\30

V_{x}=26\times 0.866

V_{x}=22.51\frac{m}{s}

V_{x}=23\frac{m}{s}

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A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re
Xelga [282]

Answer:

(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

(b) the kinetic energy of the bullet plus the block before the collision is 500J

(c) the kinetic energy of the bullet plus the block after the collision is 16.13J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki =  ¹/₂(0.1 x 100²) +  ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

6 0
3 years ago
a girl performed 50j of work lifting a heavy box it took her 5 seconds to lift the box what is her power
son4ous [18]

Answer:

10 W

Explanation:

Power is work over time.

P = W / t

P = 50 J / 5 s

P = 10 W

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3 years ago
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
Find the equivalent resistance.
frosja888 [35]

Answer:

18 Ω

Explanation:

As K and F are at the same voltage, we can redraw the diagram as in figure 2

Series resistances add directly, so we get figure 3

Adding parallel resistances gets us to figure 4

Now we can move two 6Ω resistances for clarification in figure 5

As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6

Add series resistances to get to figure 7

Add parallel resistances to get to figure 8

Add series resistances to get to figure 9

6 0
3 years ago
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The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?
Yakvenalex [24]

Answer:

3

Explanation:

the answer is number three

5 0
3 years ago
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