Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

An RV is traveling 60 km/h along a highway at night. A boy sitting near the driver of the RV turns a flashlight on and shines it

Varvara68 [4.7K]

Answer:

300,000,000 and the second question is the same.

Explanation:

Edge

5 0

3 years ago

What does Rockstrom mean by the need to "bend the curves"?

umka21 [38]

Answer: Rockstrom meant that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.

Explanation:

Some years ago, Johan Rockström, who was the executive director of Stockholm Environment Institute, and a professor at Stockholm University, was the head of an international team that was assembled in defining planetary boundaries.

The that was gathered to speak regarding issues that pertained to the Earth ane how it can be protected from being infiltrated ad protecting it from failing. During the conversation, Rockstrom said that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.

4 0

2 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$