Plate] term 3 quiz 1 physics grade 9

The troughs or zero amplitude of a standing wave are called ____________.

konstantin123 [22]

Each successive graph is at a later time. You can see from these graphs how the amplitude of the total electric field changes, but the positions of the crests and troughs (called antinodes) and places of zero field (called nodes) never change.!!!!!!!!!!!!!!!!!

7 0

3 years ago

You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m

Brut [27]

Answer:

a) Pi,c = 1066 kgm/s

b) Pi,b = 0.0075 kgm/s

c) ΔV = - 0.0007 m/s

d) ΔV = - 0.0008 m/s

Explanation:

Given:-

- The mass of the bicycle, mc = 12 kg

- The mass of passenger, mp = 70 kg

- The mass of the bug, mb = 5.0 g

- The initial speed of the bicycle, vpi = 13 m/s

- The initial speed of the bug, vbi = 1.5 m/s

Find:-

a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

Solution:-

- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:

Pi,c = vpi* ( mc + mp)

Pi,c = 13* ( 12+ 70 )

Pi,c = 1066 kgm/s

- The initial linear momentum (Pi,b) of the bug would be:

Pi,b = vbi*mb

Pi,b = 0.005*1.5

Pi,b = 0.0075 kgm/s

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

Pi = Pf

Pi,c + Pi,b = V*(mb + mc + mp)

Where, V : The velocity of the (bicycle, passenger and bug) after collision.

1066 + 0.0075 = V*( 0.005 + 12 + 70 )

V = 1066.0075 / 82.005

V = 12.9993 m/s

- The change in velocity is Δv = 13 - 12.9993 = - 0.00070 m/s

- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).

- We will apply the linear conservation of momentum similarly.

Pi = Pf

Pi,c + Pi,b = V*(mb + mc + mp)

1066 - 0.0075 = V*( 0.005 + 12 + 70 )

V = 1065.9925 / 82.005

V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 = -0.00088 m/s

7 0

3 years ago

Read 2 more answers

Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave r

Iteru [2.4K]

Answer:

$1.63\cdot 10^{-24} J$

Explanation:

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem, we have a microwave photon with wavelength

$\lambda=12.2 cm=0.122 m$

Substituting into the equation, we find its energy:

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J$

5 0

3 years ago

Which statement is true about the atom?

slavikrds [6]

Answer:

Its d

atome contain

negative electrons,

positive protons and uncharged neutrons.

Explanation:

7 0

2 years ago

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago