Moving current has electrical energy.

[9]

ad-work [718]

Answer:

9.66 m/s 15° with +y

2.59 m/s 75° with +y

Explanation:

Momentum is conserved in the y direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

10 m/s + 0 m/s = v₁ cos 15° + v₂ cos 75°

10 = v₁ cos 15° + v₂ cos 75°

Momentum is conserved in the x direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

0 m/s + 0 m/s = v₁ sin 15° − v₂ sin 75°

0 = v₁ sin 15° − v₂ sin 75°

v₁ sin 15° = v₂ sin 75°

v₂ = v₁ sin 15° / sin 75°

Substitute.

10 = v₁ cos 15° + (v₁ sin 15° / sin 75°) cos 75°

10 = v₁ cos 15° + v₁ sin 15° / tan 75°

10 = v₁ (cos 15° + sin 15° / tan 75°)

v₁ ≈ 9.66 m/s

v₂ ≈ 2.59 m/s

7 0

3 years ago

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

Andreas93 [3]

Answer:

Part a)

$\theta_2 = 15 degree$

Part b)

$\Delta t = 2.88 s$

Explanation:

Part a)

In order to have same range for same initial speed we can say

$R_1 = R_2$

$\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}$

so after comparing above we will have

$\theta_1 = 90 - \theta$

so we have

$75 = 90 - \theta_2$

$\theta_2 = 15 degree$

Part b)

Time of flight for the first ball is given as

$T_1 = \frac{2vsin\theta}{g}$

$T_1 = \frac{2(20)sin75}{9.81}$

$T_1 = 3.94 s$

Now for other angle of projection time is given as

$T_2 = \frac{2(20)sin15}{9.81}$

$T_2 = 1.05 s$

So here the time lag between two is given as

$\Delta t = T_1 - T_2$

$\Delta t = 3.94 - 1.05$

$\Delta t = 2.88 s$

5 0

3 years ago

Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

ipn [44]

Answer:

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence, E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

U = E_p + E_k + E_g

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0

4 years ago

Sarah is making bread. She measures 0.6 kg of hot water into a bowl. The water needs to be at a temperature of

Mrrafil [7]

Answer:

$\Delta m = 0.171\,kg$

Explanation:

The process of water adding is described by the First Law of Thermodynamics:

$m_{w,o} \cdot h_{w,o} + \Delta m \cdot h_{w} = (m_{w,o} + \Delta m)\cdot h_{w,f}$

The amount of additional mass is:

$m_{w,o}\cdot h_{w,o} - m_{w,o}\cdot h_{w,f} = \Delta m\cdot (h_{w,f}-h_{w})$

$\Delta m = \frac{m_{w,o}\cdot(h_{w,o}-h_{w,f})}{h_{w,f}-h_{w}}$

Given that water is incompressible, the equation can be further simplified:

$\Delta m = m_{w,o}\cdot \frac{c_{p,w}\cdot (T_{w,o}-T_{w,f})}{c_{p,w}\cdot (T_{w,f}-T_{w})}$

$\Delta m = m_{w,o}\cdot \left(\frac{T_{w,o}-T_{w,f}}{T_{w,f}-T_{w}} \right)$

$\Delta m = (0.6\,kg)\cdot \left(\frac{40 ^{\circ}C - 42^{\circ}C}{42^{\circ}C-49^{\circ}C} \right)$

$\Delta m = 0.171\,kg$

.

6 0

3 years ago

A particle moves along the x axis from the origin. The magnitude of the position vector at time t is

Maurinko [17]

1) The average velocity is $-2.1\cdot 10^5 m/s$

2) The instantaneous velocity is $64t-260t^3$

Explanation:

1)

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

$x(t) = 32t^2 - 65t^4$

where t is the time.

The position of the particle at time t = 6 sec is

$x(6) = 32(6)^2 - 65(6)^4=-83,088 m$

While the position at time t = 12 sec is

$x(12)=32(12)^2-65(12)^4=-1,343,232 m$

So, the displacement is

$d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m$

And therefore the average velocity is

$v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s$

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

$x(t) = 32t^2 - 65t^4$

By differentiating with respect to t, we find the velocity vector:

$v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3$

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0

4 years ago