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16 days ago

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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?

1 answer:

Softa [3K]16 days ago

7 0

The terminal velocity of the object is 12.58 m/s. Explanation: Terminal velocity occurs when the drag force equals the weight of the object. The gravitational force is calculated as mg = 80 * 9.8 = 784 N. The drag force can then be equated to solve for the terminal velocity, resulting in v = 12.58 m/s or v = -15.58 m/s (which is unfeasible). Thus, the terminal velocity of the falling object is 12.58 m/s.

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

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Answer:

The box experiences an acceleration with a magnitude of 2 m/s².

Explanation:

Details provided:

Mass of the box: $m=5.0$ kg

Force acting towards the east: $F=27$ N

Frictional force acting towards the west: $f=17$ N

Let’s denote the acceleration as $a$ m/s².

Thus, the net force acting on the box in the east direction is stated as:

$F_{net}=F-f=27-17=10\textrm{ N}$

According to Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Thus, the magnitude of the box's acceleration is determined to be 2 m/s².

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1 month ago

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Gamma rays may be used to kill pathogens in ground beef. One irradiation facility uses a 60Co source that has an activity of 1.0

ValentinkaMS [3465]

Answer:

Explanation:

$C_i=3.7\times 10^{16} \,decays/sec$

The energy from gamma rays generated by ore decay is

$(1.17+1.33)MeV=2.50MeV$

The energy from gamma rays created per second is

$(2.5\times 3.7\times 10^{16})MeV$

Activity

$1 \times 10^6c$

The total energy of gamma rays produced in one second is

$(2.5 \times 3.7 \times 10^{16} \times 10^6)MeV$

E represents the energy of gamma rays generated in one hour is

$(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV$

Gamma ray energy absorbed by meat = 30% (in 1 hour) of E is 0.3E

The dosage necessary to eliminate pathogens = 4000Gy=4000J/kg

The mass of meat that can be processed every hour is

$\frac{0.3E/hr}{4000J/kg}\\\\E=(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV\\\\=333\times 10^{18}MeV\\\\eV=1.6\times 10^{-19}\\\\E=53.3J$

The total meat produced = $\frac{0.3\times 53.3}{4000}=3.996\times 10^{-3}$

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2 months ago

A 3030 cmcm wrench is used to loosen a bolt with a force applied 0.30.3 mm from the bolt. It takes 6060 NN to loosen the bolt wh

Softa [3030]

Complete Question

A 30 cm cm wrench is employed to loosen a bolt, with force applied 0.30m from the bolt. It requires 60 N to loosen the bolt when force is applied perpendicular to the wrench. How much force would be necessary if the force were applied at a 30-degree angle from perpendicular?

Response:

The strength needed is $F_{\theta } = 69.28 \ N$

Clarification:

According to the problem, we know that

The length of the wrench is $L = 30 cm = \frac{30}{100} = 0.3 \ m$

The distance from the bolt is $d = 0.30 m$

The force necessary to loosen the bolt is $F = 60 N$

The angle of application is $\theta = 30 ^o$

Generally, the torque required for loosening the bolt is defined as

$\tau = F * d$

$\tau = 60 * 0.3$

$\tau = 18 Nm$

Now for the bolt to loosen at $\theta$ the torque at 90° must equal that at $\theta$

Thus, the torque at $\theta$ is represented mathematically as

$\tau = F_{\theta }d cos \theta$

substituting values gives us

$18 = F_{\theta } * 0.3 cos (30)$

$F_{\theta } = \frac{18}{0.3 cos (30)}$

$F_{\theta } = 69.28 \ N$

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19 days ago

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kicyunya [3294]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

$r = 16m$

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$\mu_s = 0.63$

The maximum velocity can be determined using centripetal force,

$F_c = \frac{mv^2}{r}$

Should be equal to,

$\frac{mv^2}{r} = \mu_s mg$

$v = \sqrt{\mu_s gr}$

$v = \sqrt{(0.63)(9.8)(16)}$

$v = 9.93m/s$

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

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1 month ago

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The answer is

-Small f and large D.

The explanation:

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-Hence, by applying a small force to the jack, the height at which the car is elevated increases.

Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

Employing leverage or hydraulic principles, machines enhance your exerted force.

Utilizing a greater lever allows for extensive movement with minimal force, resulting in the opposite side moving shorter distances with an increased force.

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1 month ago

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