A bucket of water with weight of 50N is pulled up using a rope. The bucket is moving at

¿Cuál es el parámetro que indica la cantidad de energía liberada en un movimiento sísmico?

artcher [175]

Answer:

Explanation:

en un movimiento sísmico?

3 0

2 years ago

!!please help !!!!! I’ll give brainliest

Oxana [17]

The answer is 45 degrees

7 0

3 years ago

Read 2 more answers

Blake stands in a canoe in the middle of a lake. The canoe is stationary. Blake holds an anchor mass of 15 kg, then throws it we

Inessa05 [86]

The velocity of the canoe is 1.7 m/s.

<h3>What is momentum?</h3>

Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv

a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1

b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1

c) The final velocity of the boat is obtained from;

255 Kgms-1 = (15Kg + 135 Kg) v

v = 255 Kgms-1/(15Kg + 135 Kg)

v = 1.7 m/s

Learn more about momentum: brainly.com/question/904448

5 0

2 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

How much mass energy could be obtained from the complete conversion of a 235 g hamburger?

Radda [10]

E = mc²
E = 0.235 kg · (3×10⁸ m/s)² = 0.235 · 9×10¹⁶ kg·m/s²
E = 2.115×10¹⁶ J
The answer is d) 2.12×10¹⁶ J

8 0

3 years ago