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larisa86 [58]
3 years ago
5

What is the maximum distance that a 60.-watt motor may vertically lift a 90.-newton?weight in 7.5 seconds?a. 2.3mb. 5.0mc. 140md

. 1100m
Physics
1 answer:
Naddik [55]3 years ago
5 0

Answer:

b. 5.0 m

Explanation:

Power: Power can be defined as the rate at which work is done.

The S.I unit of power is watt (W).

P = W/t ............................................... Equation 1

Where p = power, W = work, t = time.

But

W = F×d..................................... Equation 2

Where F = force, d = maximum distance.

Substituting equation 2 into equation 1

P = F×d/t

making d the subject of the equation above,

d = Pt/F......................................................... Equation 3

<em>Given: F = 90 Newton, P = 60 watt, t = 7.5 seconds.</em>

<em>Substituting these values into equation 3,</em>

<em>d = 60×7.5/90</em>

<em>d = 5 m</em>

Thus the maximum distance is = 5 m

The right option is b. 5.0 m

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What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0
3 years ago
Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

                       v_f= 11.29 m/s

3 0
3 years ago
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
What is NOT a raw minteral used in photosyntheisis
Yuliya22 [10]
The photosynthesis equation is : 

H2O + CO2 + sunlight (energy) > C6H12O6 + O2 

So water (H2O) and carbon dioxide (CO2) are the raw materials need for photosynthesis.
6 0
3 years ago
What type or types of movement occur in a hurricane as it moves along its track?
Ede4ka [16]
The answer to your question Tropical Cyclones 
8 0
3 years ago
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