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1 year ago

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

Physics

1 answer:

fenix001 [56]1 year ago

5 0

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original VERTICAL velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t = .639 seconds to hit the ground

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s = 6.39 m

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Your image appears 3.0 m from a plane mirror. How far IS YOUR IMAGE IN RELATION TO YOU?*

Vsevolod [243]

Answer:

Being a plane mirror the Image is formed 3 metres beyond the mirror . So total distance is 3+3 = 6metres

8 0

3 years ago

Read 2 more answers

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the

ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0

3 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

Where:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

Where:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

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3 years ago

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2 years ago

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larisa86 [58]

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3 years ago

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .