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3 years ago

8.

Physics

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

a. petrification

Explanation:

tar seeps = natural trade that, because of its close proximity to the ground surface, seeps from the cracks in the Earth or between rocks forming pits or pools (tar pits)

amber = fossilized resin produced by extinct coniferous trees, typically yellow in color

mummification = a process in which the skin and flesh of a corpse can be preserved by embalming and drying

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One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0

3 years ago

A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

In-s [12.5K]

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

4 0

3 years ago

A man pulls on his dog's leash to keep him from running after a bicycle. Which term best describes this example? Select one: A.

madreJ [45]

C. Negative force. The dog isn't going to learn that way.

6 0

3 years ago

For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.

andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0

3 years ago

Please help me with this question ASAP.

victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

$\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}$

Where:

E = e.m.f. of the balance point cell

$R_{balance}$ = Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω

$R_{cell}$ = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

$\dfrac{E}{7.5} = \dfrac{9}{3}$

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

7 0

3 years ago