All categories

3 years ago

What was an ill effect of the industrial revolution?

Physics

1 answer:

Digiron [165]3 years ago

6 0

Answer:

answer is option (c) child labour

You might be interested in

A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1) . After sitting a

Alekssandra [29.7K]

Answer:

0.0443 m/s

Explanation:

$m_1$ = Mass of honeybee = 0.15 g

$m_2$ = Mass of popsicle stick = 4.75 g

$v_1$ = Velocity of honeybee

$v_2$ = Velocity of stick = 0.14 cm/s

In this system the linear momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{4.75\times 0.14}{0.15}\\\Rightarrow v_1=4.43\ m/s$

The velocity of the bee is 4.43 cm/s or 0.0443 m/s

5 0

2 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

1 year ago

Consider a ball (with moment of inertia I=2/5MR^2) which bounces elastically

Oliga [24]

vn = 10nV tanθ 7 (n even), and vn = (10n−4) V tanθ 7 (n odd).

4 0

3 years ago

Read 2 more answers

How fast must a 1000 kg car be moving to have a kinetic energy of 2.0*10^3

lawyer [7]

Kinetic Energy is defined by Ke=1/2mv^2. Plug in and solve for v.

2,000 = 1/2(1000)(v)^2
4=(v)^2
v=2 m/s

The car must move at 2 m/s to have a Ke of 2,000 Joules.

6 0

2 years ago

Which type of lens will produce a virtual image

Naily [24]

When a object is close to the converging lens it behaves like a magnifying glass. Therefore convex mirror produces a virtual image. A real image occurs when a ray converges, whereas virtual image occurs when rays only Appear to converge.

7 0

2 years ago