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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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two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters
Free_Kalibri [48]

Answer:

0 m/s²

Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

Since force, F = ma where a = acceleration and m = mass,

a = F/m.

So, since for each skater, F = 0N,

a = 0 N/m

= 0 m/s²

So, the acceleration of each skater is 0 m/s²

5 0
3 years ago
Nuclide X has a higher rate of decay than nuclide Y. Based on this information, which of the following statements must be true?
marshall27 [118]

Answer: Answer down below.

Explanation:

8 0
2 years ago
a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f
Paul [167]

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

8 0
3 years ago
Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2
Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

F_n=F-F'

F_n=37.2-22.4

F_n=14.8\ N

Work done on the wagon is given by the product of net force and displacement. It is given by :

W=F_n\times d

W=14.8\ N\times 2.5\ m

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0
3 years ago
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