Ask question

Ask question

All categories

3 years ago

9

What happens with alpha, beta, and gamma radiation?

1 answer:

tresset_1 [31]3 years ago

4 0

I'm sorry but I don't really understand the question. What is the quest actually asking???

You might be interested in

Find the intensity of a 55 dB sound given I 0=10^-12W/m^2

MakcuM [25]

Answer:

3.16 × $10^{-7}$ W/ $m^{2}$

Explanation:

β(dB)=10 × $log_{10}(\frac{I}{I_{0} })$

$I_{0}$ = $10^{-12}$ W/ $m^{2}$

β=55 dB

Therefore plugging into the equation the values,

55=10 $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

5.5= $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

$10^{5.5}$ = $\frac{I}{10^{-12} }$

316227.76× $10^{-12}$ = I

I= 3.16 × $10^{-7}$ W/ $m^{2}$

5 0

2 years ago

We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient

zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0

3 years ago

Luba_88 [7]

Answer:

Bath CD jshchdhdhfhfhhfhd jpg de f for frr for gi Jhong GO by be jr jpg be

8 0

2 years ago

Read 2 more answers

What does the ideal gas law allow a scientist to calculate that the other laws do not?

Daniel [21]

The ideal gas law.
PV=nRT
P=presure
V=volume
n=number of moles
R=Gas costant
T=temperature.

Answer: a. Number of moles.

6 0

3 years ago

Read 2 more answers

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th

lozanna [386]

Answer:

$k=105.0359\times 10^4\,W.m^{-1}.K^{-1}$

Explanation:

Given:

temperature at the hotter end, $T_H=100^{\circ}C$

temperature at the cooler end, $T_C=0^{\circ}C$

length of rod through which the heat travels, $dx=0.7\,m$

cross-sectional area of rod, $A=1.1\times 10^{-4}\,cm^2$

mass of ice melted at zero degree Celsius, $m=8.7\times 10^{-3}\,kg$

time taken for the melting of ice, $t=15\times60=900\,s$

thermal conductivity k=?

By Fourier's Law of conduction we have:

$\dot{Q}=k.A.\frac{dT}{dx}$ ......................................(1)

where:

$\dot{Q}$ =rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice, $L = 334000\,J.kg^{-1}$

$Q=m.L$

$Q=8.7\times 10^{-3}\times 334000$

$Q=2905.8\,J$

Now the heat rate:

$\dot{Q}=\frac{Q}{t}$

$\dot{Q}=\frac{2905.8}{900}$

$\dot{Q}=3.2287\,W$

Now using eq,(1)

$3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}$

$k=205.4606\,W.m^{-1}.K^{-1}$

8 0

3 years ago