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3 years ago

An object, initially at rest, moves 250 m in 17 s. What is its acceleration?

Physics

1 answer:

mash [69]3 years ago

5 0

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

250 m = (0 m/s) (17 s) + ½ a (17 s)²

a = 1.73 m/s²

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The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid

timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0

3 years ago

A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius

taurus [48]

Torque acting dowward = 6 x 0.5 = 3 Nm

Torque acting to the right = 5 x 1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2

Hope this helps

5 0

3 years ago

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A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant

ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt ..............2

so area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91 $10^{-2}$ cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 $10^{-2}$ cm

so now from equation 3 we find emf

emf = - (0.500 ) 2π(9.08 $10^{-2}$ ) 1.91 $10^{-2}$

emf = - 0.005445

and magnitude of emf = 0.005445 V

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0

3 years ago

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Fed [463]

<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Let mass of earth be M and radius of earth be r.

We have

$g=\frac{GM}{r^2}$

Now

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Mass of hypothetical planet, M' = M/2

Radius of hypothetical planet, r' = 2r

Substituting

$g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}$

Option A is the correct answer.

6 0

3 years ago

What is the answer for the question -6 + (-8) =

Vaselesa [24]

The answer is negative 14

6 0

3 years ago

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