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3 years ago

9

Determine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equat

ion: W or Fg = mg)

1 answer:

loris [4]3 years ago

5 0

Answer:

15.3 sorry if i got it wrong

Explanation:

You might be interested in

Why is the distance between the sun and Neptune dim light years

djyliett [7]

The average distance from the Sun to Neptune is about 2.795 billion miles.

That's roughly 0.00048 of a light year .

3 0

3 years ago

Read 2 more answers

A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is

Kruka [31]

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

8 0

3 years ago

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

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

$cos(\alpha) = 80/320 = 0.25$

$\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0$ relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

$320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h$

7 0

3 years ago

Read 2 more answers

a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f

Paul [167]

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

8 0

3 years ago