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2 years ago

I need it in the next hour or so!

Physics

1 answer:

PSYCHO15rus [73]2 years ago

3 0

The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is

∑ F = F[a] - F[f] - F[air] = ma

3100 N - 200 N - F[air] = (650 kg) (3 m/s²)

Solve for F[air] :

F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)

F[air] = 3100 N - 200 N - 1950 N

F[air] = 950 N

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Why do clouds tend to form over land with a sea breeze and over water with a land breeze?

Masja [62]

Because the waves in the water with the fan like system.

6 0

3 years ago

Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting

loris [4]

Answer:

0.08735 kgm²

Explanation:

m = Mass of lower leg = 5 kg

L = Length of leg = 18 cm

g = Acceleration due to gravity = 9.81 m/s²

f = Frequency = 1.6 Hz

I = Moment of inertia

Time period is given by

$T=2\pi\sqrt{\dfrac{I}{mgL}}$

Also

$T=\dfrac{1}{f}$

So,

$I=\dfrac{mgL}{(2\pi f)^2}\\\Rightarrow I=\dfrac{5\times 9.81\times 0.18}{(2\pi 1.6)^2}\\\Rightarrow I=0.08735\ kgm^2$

The moment of inertia of the lower leg is 0.08735 kgm²

8 0

4 years ago

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

$\frac{a + b}{20} = \frac{a}{7}$

$a = \frac{7}{13}b$

Differentiating the above eqn w.r.t t:

$\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}$

Now, we know that:

Rate = $\frac{db}{dt} = 5\ ft/s$

Thus

$\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s$

5 0

3 years ago

Two measurments that are the same as each other are called_____?<br><br><br> What are they called???

Softa [21]

Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. Precision is independent of accuracy.

8 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago