During intercourse, Lydia constantly worries whether her partner thinks she is

A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How

alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

$\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2$

(a)

The final angular speed is given as;

$\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s$

(b) the time taken to turn 5.5 revolutions is given as

$\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s$

3 0

2 years ago

The two equal strong kids are having a tug a war. What do you expect to happen to the ball in this situation

borishaifa [10]

Answer:as per as Newtons second law, The forces exerted on the rope create tension.

As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.

Hope this helps.. :)

3 0

3 years ago

A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that

NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

$\tau=NIAB\sin\theta$

B is magnetic field

$B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T$

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0

3 years ago

What os the term for the smallest bit into which chemical substance can be divided and still have the properties of that substan

Ostrovityanka [42]

That's a molecule of the substance. You can break the molecule down further, into the atoms that make it up, but those don't have the properties of the original 'compound'.

Here's an example:

-- Sodium is a soft, slippery metal, that explodes when water touches it.

-- Chlorine is a poisonous green gas.

When an atom of Sodium and an atom of Chlorine combine, they make one molecule of a substance called "Sodium Chloride". That's SALT ! It isn't green, it isn't a gas, it isn't poisonous, it isn't soft and slippery, and it doesn't explode when water touches it.

3 0

3 years ago

E)

guapka [62]

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :

$P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa$

So, the pressure is 32666.66 Pa.

7 0

3 years ago