All categories

3 years ago

An 81.5-kg man stands on a horizontal surface.

Physics

1 answer:

OLga [1]3 years ago

7 0

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

$P=\dfrac{F}{A}$

$P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2$

P=181.11 x 10² N/m²

You might be interested in

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter

Alex787 [66]

Answer:

d₂=232.8 m

Explanation:

We know that formula for resolution given as

θres = 1.22 λ/d

Where

λ=wavelength

d=diameter

Given that resolution should be same.So we can say that

1.22 λ₁/d₁= 1.22 λ₂/d₂

λ₁/d₁= λ₂/d₂

$d_2=\dfrac{\lambda _2}{\lambda _1}d_1$

Given that

λ₂=582 nm

λ₁=2 cm

d₁=8000 km

$d_2=\dfrac{582\times 10^{-9}\times 8000\times10^3}{0.02}\ m$

d₂=232.8 m

8 0

3 years ago

Which of the following microscope parts should routinely be adjusted to control the light source and provide optimal illuminatio

yKpoI14uk [10]

Here is the answer. The microscope parts that should routinely be adjusted to control the light source and provide optimal illumination of the specimen are the following: <span>light source; condenser; specimen; objective lens; ocular lens. Hope this answers your question.</span>

3 0

3 years ago

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle

Vinil7 [7]

Answer:

1/9 of that just outside the smaller sphere

Explanation:

The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:

$E=k\frac{Q}{r^2}$