How much time is needed for a car to accelerate from 2.0 m/s to a speed of 6.0 m/s if its acceleration is 10 m/s^2

A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

3 years ago

While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for

Nimfa-mama [501]

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

5 0

3 years ago

It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?

Anon25 [30]

Explanation:

F = ma

85 N = m (15 m/s²)

m ≈ 5.7 kg

6 0

3 years ago

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

So

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

So

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

5 0

3 years ago

If the moon orbited twice as far from earth how far would it "fall" each second?

Kamila [148]

I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
-------------------------
Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
-------------------------

Kepler's 3rd law says:

   (square of the orbital period) / (cube of the orbital radius) = constant

   T₀² = K R₀³

I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.

   new-T² = K(2 R₀)³

   new-T² = 8 K (R₀)³ = 8 old-T₀²

  <u> new-T = √8 old-T</u> <=== that's what I was after

I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now √8 months.

To put a somewhat sharper point on it, the moon's period of revolution
changes from 27.322 days to 27.322√8 = 77.278 days (rounded) .

Using 385,000 km for the moon's current average distance, the current orbital speed is
   (2π x 385,000 km) / (27.322 days) = 1,024.7 m/s
(One online source says 1.023 km, so we're not doing too badly so far.)

================================================

I'm such a dummy. I don't need to go through all of that.

If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.

That's 120.86 times the Earth's radius of 6,371 km.

So the acceleration of gravity out there would be

   (1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.

   new-G = 0.000671 m/s²

Distance a dropped object falls = 1/2 g t²

   In the first second, that's 1/2 g (1)² = 1/2 g

For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.

Distance an object falls in the 1st second =

   1/2 G = 0.000336 m/s = <em>0.336 millimeter per second</em>

I estimate the probability of a mistake somewhere during this process
at approx 99.99% . But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.

5 0

4 years ago