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3 years ago

Explain why the open umbrella hit the ground after the closed umbrella. *

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

An open umbrella would not hit the ground in front of the closed one.

Explanation:

The closed umbrella would impact first because it did not have as much air pressure. The open one has air that presses against it and makes it reach where the open umbrella floats but falls slowly

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Andrew gets on his bike and begins riding.

crimeas [40]

Answer:

Push

Explanation:

Because the back tyre pushes you towards the road.

hope this helps

6 0

3 years ago

Electronegativity affects the strength of intermolecular forces. A very electronegative atom will create a stronger negative cha

ANEK [815]

Answer:

Electronegative atom will create a stronger negative charge is true.

Explanation:

Given that,

Electronegativity affects the strength of intermolecular forces.

We know that,

Electronegativity :

Electronegativity is a ability of an atom, shared electron to itself.

In periodic table,

When we move left to right in period then the electronegativity increases.

When we move up to down in a group then the electronegativity decreases.

We know that,

The intermolecular forces are dipole-dipole interactions which is the strongest intermolecular force of attraction.

If the electronegativity of atom is more, then the atom gain negative charge.

When the attraction force is more in atoms then the electronegative will be more.

Hence, Electronegative atom will create a stronger negative charge is true.

6 0

4 years ago

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY

creativ13 [48]

B) moves with a constant speed until hitting the other end.

5 0

3 years ago

A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o

IrinaK [193]

Answer:

a. $6.032\times10^{-6}C/m^2$

b. $6.816\times10^5N/C$

Explanation:

#Apply surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as $\sigma$

$\sigma=\frac{Q}{4\pi r_{out}^2}$ , and the strength of the electric field outside the sphere $E=\frac{\sigma _{new}}{\epsilon _o}$

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

$\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2$ #surface charge outside sphere.

$\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2$

Hence, the new charge density on the outside of the sphere is $6.032\times10^{-6}C/m^2$

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be $6.032\times10^{-6}C/m^2$ ,

$E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C$

Hence, the strength of the electric field just outside the sphere is $6.816\times10^5N/C$

5 0

3 years ago

A 10-kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. what is the change in mechanica

likoan [24]

The change in mechanical energy caused by the dissipative resistance force is equal to, difference between the potential energy and kinetic energy of the object.

Potential energy of the object, P.E = mgh

m is mass of the object = 10 kg

g is acceleration due to gravity = 9.8 m/s²

h= height from which it is dropped =50 m

Substituting the value we get,

P.E = 10×9.8×50 = 4900 J

Kinetic energy of the object, K.E = $\frac{1}{2}mv^{2}$

v is the velocity of the object = 26 m/s²

K.E = (1/2)×10×(26)²

= 3380 J

Change in mechanical energy caused by dissipative force = P.E ₋ K.E

= 4900 ₋ 3380 = 1520 J

4 0

3 years ago