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Gennadij [26K]
3 years ago
14

PLEASE HELP ASAP!!!!

Physics
1 answer:
likoan [24]3 years ago
6 0

Answer: the first is incorrect the second is correct the third is correct

Explanation: the fourth is correct the fifth is incorrect

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How do you know whether to put a nine or a 10 in front of the number on the weather station
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If the air pressure on the station model is 500 or more, place a 9 in front of this number. If the pressure number on the station is less than 500 add a 10 in front of this number. 

Good luck :)
7 0
3 years ago
A cup sits on a table. Due to its position, the potential energy of the cup is 3.00 joules. Ignoring frictional effects, if the
Butoxors [25]
Due to conservation of energy, half way the potential energy will be 1.5J so the remaining 1.5J is kinetic energy.
6 0
4 years ago
Read 2 more answers
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different
kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

                           P = P_a* ( 1 - βh/T_a)^(g / R*β)

                          (g / R*β) = 32.174 / 1716*0.0035 = 5.252

                            P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

                            P = 1240 lb/ft^2

- The air density method which is expressed as:

                            P = P_a - γ*h

                            P = 2116.2 - 0.07647*14,110

                            P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

                            P = P_a* e^ ( -g*h / R*T_a )

                            P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

                            P = 1270 lb/ft^2

6 0
3 years ago
Read 2 more answers
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