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mezya [45]
3 years ago
13

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t

he interface (Fig. E12.33). The density of the oil is 790 kg>m3 . (a) What is the gauge pressure at the upper face of the block
Physics
1 answer:
bezimeni [28]3 years ago
4 0

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m

density ρ = 790 kg/m³

Using expression for the gauged pressure;

p-p₀ = ρgh

where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.

we know that, acceleration due to gravity g = 9.8 m/s²

so we substitute

p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m

= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

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Question 13 of 20
Dominik [7]

Answer:

B. It has a central nucleus composed of 29 protons and 35 neutrons,

surrounded by an electron cloud containing 29 electrons.

Explanation:

Protons and neutrons are the only subatomic particles with mass, and they are located in the nucleus. If this atom has an atomic number of 29 and is copper, it must have 29 protons (protons define which element is being observed). This means all remaining mass is from neutrons. 64-29 = 35.

Electrons have no mass and orbit the nucleus in the electron cloud. Since this copper atom is neutral (we are not told it has a charge), there must be an equal number of protons and electrons.

6 0
3 years ago
Read 2 more answers
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?
cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

3 0
3 years ago
Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 2 sin2(t), y = 2 cos2(
Tanya [424]

The distance traveled by the particle at the given time interval is 0.28 m.

<h3>Position of the particle at time, t = 0</h3>

The position of the particle at the given time is calculated as follows;

x = 2 sin2(t)

y = 2 cos2(t)

x(0) = 2 sin2(0) = 0

y(0) = 2 cos2(0) = 2(1) = 2

<h3>Position of the particle at time, t = 4</h3>

x = 2 sin2(t)

y = 2 cos2(t)

x(4) = 2 sin2(4) = 0.28

y(4) = 2 cos2(4) = 2(1) = 1.98

<h3>Distance traveled by the particle at the given time interval</h3>

d = √[(x₄ - x₀)² + (y₄ - y₀)²]

d =  √[(0.28 - 0)² + (1.98 - 2)²]

d = 0.28 m

Thus, the distance traveled by the particle at the given time interval is 0.28 m.

Learn more about distance here: brainly.com/question/23848540

#SPJ1

7 0
1 year ago
A stone dropped from the top of a building reaches the ground with a velocity of 49ms¹. If the acceleration due to gravity is 9.
bezimeni [28]

Explanation:

Given:

v₀ = 0 m/s

v = 49 m/s

a = 9.8 m/s²

Find: t

v = at + v₀

49 m/s = (9.8 m/s²) t + 0 m/s

t = 5 s

6 0
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