Ask question

Ask question

All categories

pickupchik [31]

2 years ago

6

Arsene wenger pushes a box with a force of 60N over a distance 10m calculate the work done

1 answer:

alekssr [168]2 years ago

8 0

Answer:

600J

Step-by-step explanation:

work done = force x distance

= 60 x 10

= 600J

You might be interested in

Trina collects eggs from her hens.

s344n2d4d5 [400]

Answer:

56% of her eggs were medium in may

Step-by-step explanation:

if 20% of the eggs she collected in May were small, 24% of the eggs she collected in May were large. All you have to do is add them up which is 44% and minus it by 100% which is 56%

~Batmans wife dun dun duuuun

4 0

2 years ago

Let f(x) = 2x^2 + 7x - 5 and g (x) = -8x^2 -3x + 5 what is g(x) + f(x)

babunello [35]

2x^2+7x-5+(-8x^2-3x+5)=-6x^2+4x.

6 0

3 years ago

a computer technician charges $32.50 per hour plus a $75.00 service charge. Your father's firm hires him to hook up his company'

nika2105 [10]

32.50h+75.00= 107.50h I don't know if its right or it could be 32.50h+75.00= b

3 0

3 years ago

1. Triangle XYZ has vertices at X (2, -1), Y (-4, -1), and Z (2, 2). The triangle will be dilated by a scale factor of 4. What w

sweet-ann [11.9K]

Answer:

Can’t right. Now

Step-by-step explanation:

8 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago