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Ludmilka [50]
3 years ago
10

A building contractor says that a certain building foundation should support a pressure greater than 13,000 pa if the area of th

e foundation is 130 m^2 what is the max weight the building can be?
a-4.22 * 10^5 N
b-8.14 * 10^5 N
c-9.91 * 10^5 N
d-1.69 * 10^6 N
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

d. 1.69 * 10^6 N

Explanation:

Pressure is defined as force divided by area.

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Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha
Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = \frac{Qa + Qb}{2}

charge on each sphere =  \frac{-6 + 2}{2} = \frac{-4}{2} = <em>-2 units of charge</em>

8 0
3 years ago
The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b
Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, \mu_o=4\pi\times 10^{-7}\ N.s^2/C^2

Let \epsilon_o is the permittivity of free space. The relation between \mu_o,\epsilon_o\ and\ c is given by :

c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}

\epsilon_o=\dfrac{1}{c^2u_o}

\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}

\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2

Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
A 7.00 g bullet, when fired from a gun into a 1.00 kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm b
Sauron [17]

Answer:

Explanation:

ooooooooooooooo

5 0
3 years ago
a cement block accidentally falls from rest from the ledge of a 80.6-m-high building. When the block is 10.8 m above the ground,
fomenos

Answer:

0.229 seconds

Explanation:

Given:

y₀ = 80.6 m

v₀ = 0 m/s

a = -9.8 m/s²

We need to find the difference in times when y = 10.8 m and y = 2.10 m.

When y = 10.8 m:

y = y₀ + v₀ t + ½ at²

10.8 = 80.6 + (0) t + ½ (-9.8) t²

10.8 = 80.6 − 4.9 t²

4.9 t² = 69.8

t = 3.774

When y = 2.10 m:

y = y₀ + v₀ t + ½ at²

2.10 = 80.6 + (0) t + ½ (-9.8) t²

2.10 = 80.6 − 4.9 t²

4.9 t² = 78.5

t = 4.003

The difference is:

4.003 − 3.774 = 0.229

The man has 0.229 seconds to get out of the way.

5 0
3 years ago
How much energy is needed to raise the temperature of a 55g sample of aluminum from 22.4 to 94.6?
BaLLatris [955]
The heat needed is given by Mcθ , where m is the mass in Kg, c is the heat capacity of aluminium, and θ is the change in temperature. 
Specific heat capacity of aluminium is 0.9 j/g°c
thus; Heat = 55 × 0.9 × 72.2
                 = 3573.9 Joules or 3.574 kJ
3 0
4 years ago
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