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3 years ago

A 20 kg bike accelerates at 10 m/s2. With what force was the person pedaling if friction

Physics

2 answers:

denpristay [2]3 years ago

8 0

Answer:200N

Explanation:

Molodets [167]3 years ago

4 0

Look it up plz cuz like I’m slow so

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Protons and neutrons grouped in a specific pattern

alexgriva [62]

Answer b protons and electrons

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3 years ago

1.) What causes ocean tides

BigorU [14]

Answer: Tides are periodic rises and falls of large bodies of water. Tides are caused by gravitational interaction between the earth and the moon. The gravitational attraction of the moon causes the oceans to bulge out in the direction of the moon.

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3 years ago

Read 2 more answers

A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10-12 W/m2.

ahrayia [7]

Answer:

0.0018 W/m²

Explanation:

Power and intensity are related as:

$I=\frac{P}{4\pi r^2}$

P= 20.0 W (given)

r = 30.0 m (given)

$I=\frac{20.0}{4\pi(30.0)^2}=0.0018 W/m^2$

Intensity in decibels:

$I(dB)=10log\frac{I}{I_o}\\=10log\frac{0.0018}{10^{-12}}=92.5 dB$

7 0

3 years ago

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

3 years ago

An electron is moving horizontally east in an electric field that points vertically downward. The electric force on the electron

saveliy_v [14]

Answer:

d. )directed upward.

Explanation:

As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.

This is because the electric field has the same direction that the force on a positive test charge at the same point.

As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.

So, the statement d. is the one that results to be true.

7 0

3 years ago