Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.

where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
![A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }](https://tex.z-dn.net/?f=A%20%3D%204%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B100%5E%7B2%7D%20cm%5E%7B2%7D%20%7D)
A = 0.0004 [m²]
Also, the weight should be calculated as follows:

where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
![w = 60*9.81\\w = 588.6[N]](https://tex.z-dn.net/?f=w%20%3D%2060%2A9.81%5C%5Cw%20%3D%20588.6%5BN%5D)
And the pressure:
![P=588.6/0.0004\\P=1471500 [Pa]](https://tex.z-dn.net/?f=P%3D588.6%2F0.0004%5C%5CP%3D1471500%20%5BPa%5D)
Because 1 [Pa] = 1 [N/m²]
Answer: The scientist gives up and starts an investigation on a new topic.
Explanation:
The data is altered so that it supports the original hypothesized. The data is then altered so that it supports the original hypothesis.
Answer:
1.4 * 10 ^-1 Ω
Explanation:
Hi,
For this question, we gotta use the formula
R = pL/A
p = The resistivity of your material at 20°C
L = length of the wire
A = cross-sectional area
The resistivity of tungsten is 5.60 * 10^-8 at 20°C
By plugging the values, we get:
R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω
Answer:
Explanation:
During an energy transfer, the collision loss for an electron can be determined by using the formula:

However; from the total stopping power & power loss of the electron;

where;
Z = atomic no. for lead = 82
E = 1.9 MeV
∴
radiational energy loss = collisional energy loss 
= 0.19475
b)
Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.
Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.
Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.
Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.
d. 49.0 m/s
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