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GuDViN [60]
2 years ago
13

PLEASE HELP MEE PLEASE I BEG

Physics
1 answer:
TEA [102]2 years ago
8 0

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

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LO NECESITO PARA HOY URGENTE!! 2. Exprese en metros las siguientes longitudes a) 48,9 km b) 36,875 cm c) 756,34 hm d) 9876 mm
Vesnalui [34]

Answer:

a)48900 metros

b)0.36875 metros

c)75634 metros

d)9.876 metros

Explanation:

Hola, para resolver debemos convertir unidades utilizando equivalencias

a) 48.9 km  

1 kilometro = 1000 metros

48.9 x 1000 = 48900 metros

b) 36.875 cm  

1 centímetro =0.01 metros

36.875 x 0.01 = 0.36875 metros

c) 756,34 hm

1 hectómetro= 100 metros

756.34 x 100 = 75634 metros

d) 9876 mm

1 milímetro = 0.001 m

9876 x 0.001 = 9.876 metros

4 0
3 years ago
Suppose a large housefly 3.0 m away from you makes sound with an intensity level of 40.0 dB. What would be the sound intensity l
melisa1 [442]

Answer:

Explanation:

Let the intensity of the noise be represented by I

Given that

40dB = 10 log 10 ⁡ ( I /I•) ........ 1

I• is the lowest or threshold intensity of sound made.

I represents the intensity of the sound/ noise

The intensity of noise of 1000flies will be

β = 10 log 10 ⁡(1000I/I•)

Open up the bracket

β = 10 log 10(1000)+ 10 log 10(I/I•)

10 log 10(10^3)+10 log 10(I/I•)

3×10(10 log 10) +10 log 10(I/I•)

Recall, 10 log 10 = 1

30×1 + 10 log 10(I/I•).........2

Put equation 1 into 2

β =30+40

= 70db

5 0
2 years ago
A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent hea
Sunny_sXe [5.5K]

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

7 0
3 years ago
What causes a satellite to move tangentially to a circular path?
yawa3891 [41]

Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

So, we can say gravity is responsible for the motion of the satellite in a circular path.

5 0
3 years ago
Read 2 more answers
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

7 0
3 years ago
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