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Tresset [83]
2 years ago
8

The weight of object is more in the poles than the equtor of the earth.why?​

Physics
1 answer:
vagabundo [1.1K]2 years ago
4 0

hope it helps

plz plz mark brainliest

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A cat walks along a plank with mass M= 7.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
solong [7]

Answer:

Before the plank will tip cat will walk 1.652 m

Explanation:

Mass of the cat along with plank m_1=7kg

Center of mass of the plank d_1=0.850m

Mass of cat m_2=3.6kg

We have to find how far right of sawhorse B.

Plank will tip when weight of the cat about B is greater than the torque by the weight of the plank.

Balancing the the torque

m_1gd_1=m_1gd_2

7\times 9.8\times 0.850=3.6\times 9.8\times d_2

d_2=1.652m

5 0
3 years ago
A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have internal pressure strength of at le
maria [59]

Answer: K =24 psi

Explanation:

Given: Standard deviation =3psi

Internal pressure strength =157psi

Number of random bottle =n=64

K= 3 × square root of 64

K= 3×8=24 psi

If mean internal pressure K fall below K,

157-1.3=155.7psi

At 2%:

0.16×64 = 10.24

5 0
2 years ago
Which best describes the relationship between the direction of energy and wave motion in a transverse wave?
sammy [17]
I think the correct answer from the choices listed above is the second option. The relationship between the direction of energy and wave motion in a transverse wave would be the <span>energy direction is perpendicular to the motion of the wave. Hope this answers the question. Have a nice day.</span>
6 0
3 years ago
Read 2 more answers
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Which ion will act as a weak base in water?
GuDViN [60]

Answer:

An example of a weak base is ammonia. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. The position of equilibrium varies from base to base when a weak base reacts with water. The further to the left it is, the weaker the base.

Explanation:

I hope this helps!!

4 0
2 years ago
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