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3 years ago

Please need help on this not too sure on this

Physics

2 answers:

Bezzdna [24]3 years ago

8 0

The last one because if the lady bug is small it will look big looking through the magnify glass

n200080 [17]3 years ago

3 0

Answer:

the last one

Explanation:

Because it is a magnifying glass, it magnifies the object and makes it bigger than it appears

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The layers of gas that surround the Earth

Ierofanga [76]

Answer:

Our planet is surrounded by a layer of gases called the atmosphere. ... ➢ Without our atmosphere, there would be no life on earth. ➢ Scientists divided the atmosphere into four layers according to temperature: troposphere, stratosphere, mesosphere, and thermosphere.

Explanation:

8 0

3 years ago

Read 2 more answers

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 large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the

exis [7]

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

7 0

3 years ago

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List down atleast five activities where friction gave you an undesirable experience

Step2247 [10]

Answer:

Rug burn, Indian burn done to you by a friend, friction from the road causes your car to accelerate at a slower rate, The cylinder heads in an engine, When trying to move a heavy object across a rough surface

Explanation:

7 0

3 years ago

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Suppose an electron is trapped within a small region and the uncertainty in its position is 24. 0 x 10-15 m. what is the minimum

sergeinik [125]

The minimum uncertainty in the electron's momentum is

$Δp = 2.2822 \times 10 {}^{ - 20} kg/ms$