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3 years ago

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How can you increase the pressure on an object?

1 answer:

Cerrena [4.2K]3 years ago

8 0

Increase the force on the object

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

• How does wind shape Earth’s surface?

ehidna [41]

Lamina and turbulent flow

Explanation:

mentioning about lamina and turbulent flow we could say that both form in different period of time

6 0

3 years ago

Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei

mars1129 [50]

Explanation:

The given data is as follows.

Fluid is water so, density $\rho = 1000 kg/m^{3}$

Weight flow rate = 500 lbf/s = 2224.11 N/sec

Cross-sectional area (A) = $\pi \times \frac{30}{2} \times \frac{22}{2}$

= 0.05184 $m^{2}$

Hence, weight flow rate will be given as follows.

w = $\rho \times g \times A \times V$

2224.11 N/sec = $\rho \times g \times A \times V$

V = $\frac{2224.11}{1000 \times 9.81 \times 0.05184}$ m/s

= 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0

3 years ago

Read 2 more answers

Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann

chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

$v_{mp}=\sqrt{\frac{2K_bT}{m}}$

$K_b$ = Boltzmann’s constant = $1.38066\times 10^{-23} J/K$

T = temperature of the gas

m = mass of the gas particle.

Given, m = $6.65\times 10^{-27} kg$

T = 6.4 K

Substituting all the given values :

$v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}$

$v_{mp}=162.78 m/s$

162.78 m/s is the most probable speed of a helium atom.

4 0

3 years ago

If the new moon happens on January 15th, what shape will it be on February 6th?

Jobisdone [24]

-- From January 15 to February 6 is a period of 22 days.

-- The period of the full cycle of moon phases is 29.53 days.

-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.

-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at <em>Third Quarter</em>. That's the <em>left half of a disk </em>(viewed from the northern hemisphere).

3 0

3 years ago