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2 years ago

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Q.1 Lorna built the circuit diagram below. All the bulbs are identical. (a) Complete the table below by writing on or off for ea

ch bulb.

1 answer:

irga5000 [103]2 years ago

8 0

Answer:

on and off

Explanation:

if there are switches, it can change if the electricity can get to the bulb or not. if it appears that there is no pathway for the electricity to get to the light bulb, it is of, if there is a pathway, its on

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the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

dem82 [27]

Answer:

<em>Time = 5 seconds</em>

<em>Distance = 50 meters</em>

Explanation:

<u>Constantly Accelerated Motion</u>

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

$v_f=v_o+a.t$

And the distance s is

$\displaystyle s=v_o.t+\frac{gt^2}{2}$

Given the object starts from rest, vo=0 and vf=20 m/s at $a=4\ m/s^2$ . We compute t

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}$

$\boxed{t=5\ sec}$

Now we compute s

$\displaystyle s=0+\frac{4\times 5^2}{2}$

$\boxed{s=50\ m}$

5 0

2 years ago

Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.

melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

$P.E_{1} = mgz_{1}$

$P.E_{1} = (\rho V)gz_{1}$

= $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

= 1164000 J

or, = 1164 kJ (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

$\Delta P.E = mg(z_{2} - z_{1})$

= $\rho \times V \times g (z_{2} - z_{1})$

= $1000 \times 3 \times 9.7 (10 - 40)m$

= -873000 J

or, = -873 kJ

Thus, we can conclude that change in gravitational potential energy is -873 kJ.

4 0

2 years ago

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea

motikmotik

Explanation:

Work done by gravity is given by the formula,

W = $\rho A (h_{1} - h)g (h - h_{2})$ ......... (1)

It is known that when levels are same then height of the liquid is as follows.

h = $\frac{h_{1} + h_{2}}{2}$ ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

W = $\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})$

= $\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})$

= 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0

3 years ago

A student walks the hallway for 25 m stops to talk and continues down the hallway another 10 m what is the distance and displace

brilliants [131]

Answer:

Distance = displacement = 35m

Explanation:

The distance of the student is how far he has gone.

Distance = 25m + 10m

Distance = 35m

Displacement is the distance specified in specific direction. Since the student walk in the sane direction, thence the displacement is also 35m

3 0

3 years ago

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0

3 years ago