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svp [43]
3 years ago
10

Help me please I only have a few minutes to get this doneplease and thank q

Physics
1 answer:
DIA [1.3K]3 years ago
6 0
A: particles are more spread out in gas
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A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
A block slides down an incline plane that makes a 30 degree angle with the
Alona [7]

Hi there!

We know that:

Force due to gravity = Mgsinθ

Force due to friction = μMgcosθ

Let the positive direction be directed in the direction of the block's acceleration, which is downward.

Thus:

ΣF = Mgsinθ - μMgcosθ

Solving for acceleration requires diving all terms by the mass, so:

a = gsinθ - μgcosθ

Substitute in given values. (g = 9.8 m/s²)

a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²

7 0
3 years ago
A bomb of mass 4 kg, initially at rest, explodes breaking into two fragments of 1 kg and 3 kg. Which one of the following statem
Kazeer [188]

Answer: The 1 kg fragment will have three times the speed of the 3kg fragment.

Explanation:Here for the bomb, its chemical energy gets converted into the mechanical energy.

According to the law of conservation of momentum, the two bodies will have equal momentum and to satisfy this condition the lighter mass will have the higher velocity.

∵ momentum, p = mass × velocity

∴The 1 kg fragment will have three times the speed of the 3kg fragment.

6 0
3 years ago
What are isotopes. <br> Please answer soon
sweet-ann [11.9K]

each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element.

5 0
3 years ago
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