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CaHeK987 [17]
3 years ago
9

A 0.5 kg mass moving East with a velocity of 8 m/s is acted upon by a force of 20 N for 2.0 sec.

Physics
1 answer:
Andrej [43]3 years ago
7 0

Answer:b

Explanation:

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An ice skater is moving in a circle at a constant speed. Which of the following best explains the forces acting on the ice skate
dangina [55]
B because the skater is constantly moving in the same speed to go in a circle .
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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
BartSMP [9]

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
A child pushes a 75 N toy car across the floor. What is the mass of the car?
GaryK [48]

Answer:

7.6 kg

Explanation:

w=75N

w=mg

m=w÷g

m=75÷9.8

m=7.6kg

8 0
3 years ago
A 357 kg box is pulled 8.00 m up a 30° frictionless, inclined plane by an external force of 4925 N that acts parallel to the pla
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I'm so tremble right now
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3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
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