GIVING THE BRAINEST AND 15 POINTS

Forms of energy! Write a brief summary about each form of energy below.

stepan [7]

Answer:

Hope this helps!! :D

Explanation:

Mechanical energy is power that an object gets from its position and motion. the energy released during nuclear fission or fusion, typically to make electricity.

Thermal energy is the energy given off by hot things like fire.

Chemical energy is energy stored in the bonds of chemical compounds.

Electromagnetic energy is a form of energy that is reflected or emitted from objects in the form of electrical and magnetic waves that can travel.

Electrical energy is energy derived from electric potential energy or kinetic energy.

Nuclear energy is the energy released during nuclear fission or fusion, typically used to generate electricity.

4 0

3 years ago

PLEASE HELP ME!! a police car moves at a speed of 30m / s, the driver is at a single-note frequency of 650Hz It is heard. One of

irina [24]

Explanation:

r u trapped by the traffic ..

7 0

3 years ago

a car moving on a road passes by kilometer 218 at 10:15 am and milestone 236 at 10:30 am. determine an average scalar speed in k

Leno4ka [110]

Answer:

v?

Explanation:

v = ∆s / ∆t

v =?

∆s = 236 - 218 = 18km

=t = 10: 30 - 10:15 = 15 minutes to hour 15/60 = 0.25h

v = 18 / 0.25

v = 72km / h

speed of 72km / h

5 0

3 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

KatRina [158]

Answer:

T₂ = 123.9 N, θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

T₁ₓ - $T_{2x}$ = 0

T_{2x}= T₁ₓ

Y axis y

T_{1y} + T_{2y} - 200N = 0

T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

sin 60 = T_{1y} / T₁

cos 60 = t₁ₓ / T₁

T_{1y} = T₁ sin 60

T1x = T₁ cos 60

T_{1y}y = 100 sin 60 = 86.6 N

T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

T_{2y} = T₂ sin θ

T₂ₓ = T₂ cos θ

we substitute

T₂ sin θ= 200 - 86.6 = 113.4

T₂ cos θ = 50 (1)

to solve the system we divide the two equations

tan θ = 113.4 / 50

θ = tan⁻¹ 2,268

θ = 66.2º

we caption in equation 1

T₂ cos 66.2 = 50

T₂ = 50 / cos 66.2

T₂ = 123.9 N

8 0

3 years ago