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Elena-2011 [213]
2 years ago
5

What forms when air masses with different temperatures meet?

Physics
2 answers:
iVinArrow [24]2 years ago
8 0

Answer:

fronts

Explanation:

WARRIOR [948]2 years ago
7 0
Clouds and fog and it may rain or snow
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If you place a piece of paper containing a small letter "d" on the stage, what will the image look like under the microscope?
nadezda [96]

Answer:

it will be

B.

p

Explanation:

7 0
2 years ago
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A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.4. The mass of th
Sloan [31]

Answer:

Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]

Explanation:

To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.

The gravitational force is equal to:

Fg = (10 * 9.81) = 98.1 [N]

Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.

N - Fg = 0

N = Fg

N = 98.1 [N]

The friction force is defined as the product of normal force by the coefficient of friction.

Ff = N * μ

Ff = 98.1 * 0.4

Ff = 39.24 [N]

By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.

60 - Ff = m * a

60 - 39.24 = 10 * a

a = 2.076[m/^2]

6 0
2 years ago
What are the 3 formulas you can use for vertical motion for a projectile?
klio [65]
I hope this can help you ask me if you need help again

4 0
3 years ago
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
HELPP !! pleaseeeeeeeee!!!!!!!!
sertanlavr [38]

Answer:

Atomic name is your answer.

5 0
3 years ago
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