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Nuetrik [128]
3 years ago
13

Could you help me with a science question really quick?

Physics
1 answer:
Rama09 [41]3 years ago
4 0
Yea I would love to help u
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A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal.
Elza [17]

Answer:

Normal force: 167.48 N

Explanation:

  • First of all it is necessary to draw the free body diagram of the suitcase adding alll the forces stated on the question: the normal force, the friction force and pull force exterted by the woman. Additionally, we need to add the weight, the forces exerted by Earth's gravity. I attached the diagram so you can check it.
  • We need to resolve all the unknown quantities on this exercise, so we need to write down the sum of forces equations on X-Axis and Y-axis. Remember that force exerted by the woman has an angle with respect the horizontal (X-Axis),  that is to say it has force compoents on both X and Y axis.  The equations  will be equal to zero since the suitcase  is at constant speed (acceleration is zero).

        ∑F_{x}:  F{x}-20 = 0  

        ∑F_{y}: N -W+F_{y}=0

  • Our objetive is to find the value of the normal force. It means we can solve the sum of Y-axis for N. The solution would be following:

       N = W - Fy

  • Keep in mind Weight of the suitcase (W) is equal to the suitcase mass times the acceleration caused by gravity (9.81\frac{m}{s^{2}}. Furthermore, Fy can be replaced using trigonometry as Fsin(\theta) where θ is the angle above the horizontal. So the formula can be written in this way:

        N = mg -Fsin(\theta)

  • We need to find the value of θ so we can find the value of N. We can find it out solving the sum of forces on X-axis replacing <em>Fx</em> for Fcos(θ). The equation will be like this:

        Fcos(\theta) -20 = 0   ⇒   Fcos(\theta) = 20\theta                            [tex]\theta=cos^{-1}(20/F)

  • Replacing the value of F we will see θ has a value of 55.15°. Now we can use this angle to find the value of N. Replacing mass, the gravity acceleration and the angle by their respective values, we will have the following:

       N = 20 x 9.81 - 35sin(55.15)  ⇒ N = 167.48 N

7 0
3 years ago
Study the image of earths layer which statement correctly compares the thicknesses of earths layers
My name is Ann [436]

This question is incomplete because the options are missing; here is the complete question:

Study the image of the Earth's layer which statement correctly compares the thicknesses of earths layers

A. Earth’s mantle is thinner than its oceanic crust.

B. Earth’s outer core is thicker than its mantle.

C. Earth’s continental crust is thicker than its lithosphere.

D. Earth’s lithosphere is thinner than its asthenosphere.

The answer to this question is D. Earth’s lithosphere is thinner than its asthenosphere.

Explanation:

The image shows the different layers that are part of Earth, as well as the thickness of each layer, in kilometers. In this, the thickest layer is the Mantle that is almost 2900 kilometers; this is followed in thickness by the outer and the inner core.

Additionally, other layers such as the continental/oceanic crust, the asthenosphere, and the lithosphere that are near the surface are thinner. About this, it can be concluded the lithosphere is thinner than the asthenosphere because the lithosphere has a thickness of 100 km, while the asthenosphere thickness is 660km. This makes option D the correct.

4 0
3 years ago
The figure below shows a dipole. If the positive particle has a charge of 37.3 mC and the particles are 3.08 mm apart, what is t
kolbaska11 [484]

The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

<h3>Electric field of the positive particle</h3>

The electric field is calculated as follows;

E = kq/r²

where;

  • r is the distance between the charges
  • k is Coulomb's constant
  • q is magnitude of the charge

midpoint of 3.08 m, x = 1.54 mm

r(1.54 mm, 2.00 mm)

|r| = √(1.54² + 2²)

|r| = 2.52 mm

E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²

E = 5.287 X 10¹³ N/C

Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

5 0
2 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
3 years ago
What are the smallest parts that make up matter?
Alecsey [184]
The smallest parts that make up different types of matter are called atoms.
Atoms are tiny elements that are made up of literally everything on earth. <span />
3 0
4 years ago
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