1. How much power does a light bulb contain if it does 600J of work in 5 seconds?

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

Solving the equation (1) for d, we find

$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

5 0

3 years ago

Why is a shadow formed

a_sh-v [17]

Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

6 0

3 years ago

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A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

2 years ago

(1) A hydraulic press is used to lift heavy objects. It consists of a U-tube, one end having a large cross-sectional area A1, th

lana [24]

Answer:

The load that can be lifted is equal to the weight W = F2A1/A2

Explanation:

According to Pascal principle which states that the pressure applied to a liquid confined in a container will be transmitted equally to all other parts of the container.

Since pressure = Force/Area

The force F2 applied at one end of the piston will generate a pressure of F2/A2. This pressure generated will be transmitted to the other end of the piston of area A1 to lift the load through a distance.

The piston where the load is will experience an upward force F1 which is equal to Pressure × Area.

The pressure experienced by the load is applied by force F2.

Force on the load = (Pressure exerted by Force F2) × Area at the larger end A1

Force on the load = F2/A2 × A1

Since the load experiences a weight W

The weight will be equal to the force on the load which is to be lifted i.e W =Force on the load.

W = F2A1/A2

The load that can be lifted is equal to the weight W = F2A1/A2

6 0

3 years ago

Tommy was walking at a rate of 4 miles hour at noon and at 12:30 pm he was walking at a brisk rate of 6 miles hour . Two hours l

maks197457 [2]

Answer:

B) Tommy had a positive acceleration between noon and 12:30 pm.

Explanation:

Acceleration is defined as the rate of change of velocity:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

In the problem,

- At noon, Tommy is walking at a velocity of 4 mi/h

- At 12.30 pm, Tommy is walking at a velocity of 6 mi/h

- A time of half an hour (0.5 h) passed between the two moments

So Tommy's acceleration is

$a=\frac{6 mi/h-4 mi/h}{0.5 h}=4 mi/h^2$

and the acceleration is positive, since the velocity has increased.

3 0

3 years ago

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