100 J of work in 10 seconds

A tow truck is pulling 25,000 kg van with a tow hook that meets the van at an angle of 30° with the ground. How much force must

Law Incorporation [45]

Answer:

86605.08 N

Explanation:

The equation to calculate the force is:

Force = mass * acceleration

The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:

Force_x = Force * cos(30)

Then, we have that:

Force_x = mass * acceleration

Force * cos(30) = 25000 * 3

Force * 0.866 = 75000

Force = 75000 / 0.866 = 86605.08 N

3 0

4 years ago

A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie

ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, $A=7.4\ cm^2=7.4\times 10^{-4}\ m^2$

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, $B_1=0.5\ T$

Final magnetic field, $B_2=3\ T$

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_2-B_1}{t}$

$\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}$

$\epsilon=1.65\times 10^{-3}\ V$

Induced current in the loop of wire is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.65\times 10^{-3}}{2.4}$

$I=6.87\times 10^{-4}\ A$

So, the induced current in the loop of wire over this time is $6.87\times 10^{-4}\ A$ . Hence, this is the required solution.

7 0

3 years ago

3. (a) Sate any three properties of an ideal gas as assumed by the kinetic theory of gas.

Aliun [14]

Explanation:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the ...

4 0

2 years ago

El conductor de un tren que circula a 20 m/s ve un obstáculo en la vía y frena con una aceleración de 2 m/s2 hasta parar ¿cuánto

AlladinOne [14]

Velocidad inicial = 20 m/s
velocidad final = 0 m/s
aceleracion = -2 m/s^2

aceleracion = (cambio de velocidad)/(cambio de tiempo)
(cambio de tiempo)= (cambio de velocidad)/aceleracion
tiempo = (-20 m/s)/(-2 m/s^2)
= 10 segundos

x = (x(inicial)) + (v(inicial))(tiempo) + 1/2(aceleracion)(tiempo)^2
x(inicial) = 0
x = (20 m/s)(10 s) + 1/2 (-2m/s^2)(10 s)^2
x = 200 m - 100 m
x = 100 m (el espacio recorrido en los dos segundos)

espero que esto te ayude! buena suerte!

6 0

3 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges $\rm q_1$ and $\rm q_2$ which are separated by the distance $\rm d$ is given by

$\rm F = \dfrac{kq_1q_2}{d^2}.$

<em>where,</em> k is the Coulomb's constant.

For the case, when,

$\rm q_1 = Q.$
$\rm q_2 = Q.$
$\rm d=r.$
$\rm F=12.3442.$

Then, using Coulomb's law,

$\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).$

For the case, when,

$\rm q_1 = 2Q.$
$\rm q_2 = 2Q.$
$\rm d=\dfrac r2.$

Then, using Coulomb's law, the new electric force between the charges is given by,

$\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.$

8 0

3 years ago