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3 years ago

ASAP please help me

Physics

1 answer:

just olya [345]3 years ago

4 0

Answer:

1.2 s

Explanation:

We'll begin by calculating the length (i.e distance) of the ramp. This can be obtained by using pythagoras theory as illustrated below:

NOTE: Length of the ramp is the Hypothenus i.e the longest side.

Let the Lenght of the ramp be 's'. The value of x can be obtained as follow:

s² = 4² + 3²

s² = 16 + 9

s² = 25

Take the square root of both side

s = √25

s = 5 m

Thus the length of the ramp is 5 m

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

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

v² = u² + 2as

v² = 3² + (2 × 2 × 5)

v² = 9 + 20

v² = 29

Take the square root of both side

v = √29

v = 5.39 m/s

Finally, we shall determine the time taken for the ball to reach the final position. This can be obtained as follow:

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Final velocity (v) = 5.39 m/s

Time (t) =?

v = u + at

5.39 = 3 + 2t

Collect like terms

5.39 – 3 = 2t

2.39 = 2t

Divide both side by 2

t = 2.39 / 2

t = 1.2 s

Thus, it will take 1.2 s for the ball to get to the final position.

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2 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. $\omega _0=0 rad/s$

after $t=10 s$

$\omega =10 rad/s$

using $\omega =\omega _0+\alpha t$

$10=0+\alpha \cdot 10$

$\alpha =1 rad/s^2$

Revolutions turned in 2 s

$\theta =\omega _0t+\frac{\alpha t^2}{2}$

$\theta =0+\frac{1\times 2^2}{2}$

$\theta =2 rad$

To get revolution $\frac{\theta }{2\pi }$

= $\frac{2}{2\pi}=0.318\ revolutions$

3 0

3 years ago

Read 2 more answers

Which of the following distinguishes the isotope uranium-238 from the isotope uranium-235?

WITCHER [35]

Answer: A

Explanation:

Isotopes of different elements differ by the number of neutrons inside the nucleus.

8 0

2 years ago

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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav

DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

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v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s$

Time taken to reach 12% of light speed is 3673469.39 seconds

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m$

The distance it would have to travel is 6.61×10¹⁴ m

7 0

3 years ago

ASAP please help me ​

ASAP please help me