Which series of energy transfers happens when light from the sun is used to power a light bulb?

What does one prolonged blast emitted by boat a using an efficient sound producing device indicate?

Anika [276]

A single prolonged blast emitted by a boat using an efficient sound producing devices indicates that the boat is leaving the dock. It may also be used to indicate that the boat is moving and for alarming other vessels, but in this case, one prolonged blast is followed by two short blasts, in two minutes of time.

3 0

3 years ago

Am i right on this one?

NISA [10]

well if each square is 6 km, then the car DOES go 6 km, but it also moves WEST, not east. i would say that since its displacement not distance, its 2 km WEST :)

4 0

3 years ago

Usain Bolt, a Jamaican sprinter, holds the Olympic and world records for the 100-m and 200-m dash, which he

stellarik [79]

Answer: 10.36m/s

How? Just divide 200m by 19.3 and you will get how fast he ran per m/s

6 0

3 years ago

An iron anchor of density 7890.00 kg/m3 appears 299 N lighter in water than in air. (a) What is the volume of the anchor? (b) Ho

PtichkaEL [24]

Answer:

weigh is 2353.13 N

Explanation:

Given data

density = 7890.00 kg/m3

lighter = 299 N

to find out

the volume of the anchor and weigh in air

solution

from question we can say that

apparent weight = actual weight - buoyant force

we know weight = mg and buoyant force = water density × g

so volume of anchor is = actual weight - apparent weight / buoyant force

volume of anchor is = 299 / 1000 × 9.81

volume of anchor is = 0.0304791 m³

and

weight of anchor is mg

here mass m = density Fe g

density Fe = 7870 from table 14-1

so weight = 7870 × 0.0304791 × 9.81

weigh is 2353.13 N

7 0

4 years ago

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using