What will happen if you light up a candle that is in a beaker?

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b

GuDViN [60]

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

3 0

3 years ago

1) A person riding their bike on a nice cloudy day travels 50 meters in 20 seconds.

Rasek [7]

Answer:

5.59

Explanation:

50 meters in 10 seconds is 11.18, which is an easy way to remember. Just divided by 2

5 0

2 years ago

What is the density of a 150cm 3 block with a mass of 700 grams?

Vinvika [58]

Answer:

The Density of the block is 4.667g/mL

Explanation:

Given the following data;

Mass of block = 700g

Volume of block = 150cm³

Density = ?

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

$Density = \frac{mass}{volume}$

Substituting into the equation, we have;

$Density = \frac{700}{150}$

<em>Density = 4.667g/mL</em>

3 0

3 years ago

Can someone please help meee!!

Paraphin [41]

The answer is 9800 J/S

5 0

3 years ago

A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

$a=\frac{v}{t}$ Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).