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3 years ago

When a student stands on a rotating table,the frictional force exerted on the student by the table is?

Physics

1 answer:

lakkis [162]3 years ago

7 0

Answer:

Less

Explanation:

because static friction is more than rolling friction

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WhAT is the change IN THE entropy of 2.0kg of h2o molecules when transform at a constant pressure of 1 atm from water at 100 deg

Fynjy0 [20]

Answer:

The entropy change is 45.2 kJ/K.

Explanation:

mass of water at 100 C = 2 kg

Latent heat of vaporization, L = 2260 kJ/kg

Heat is

H = m L

H = 2 x 2260 = 4520 kJ

Entropy is given by

S = H/T = 4520/100 = 45.2 kJ/K

3 0

2 years ago

A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?

azamat

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

R = 2ohms

3 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius..

Troyanec [42]

Answer:

Answer:

= 338.2 m/s

Explanation:

vs= 331 + T (0.6m/s)

= 331 + 12 °C (0.6m/s)

= 331 + 7.2m/s

= 338.2 m/s

Explanation:

6 0

3 years ago

Locate the mode of 12, 3, 5, 17, 3, 18, 5, 11, 11, 15, 3, 9, 3.

zimovet [89]

Answer:
Mode = 3 because it is listed 4 times

8 0

3 years ago