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3 years ago

Please need help with this

Physics

2 answers:

Stella [2.4K]3 years ago

4 0

Answer:

D. The objects particles move faster

ozzi3 years ago

3 0

Answer:

D. The objects particles move faster

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Six dogs pull a two-person sled with a total mass of 220 kg. The coefficient of kinetic friction between the sled and the snow i

irina [24]

Answer:

a) 2457J

b) 558W

c) 337N

Explanation:

Assuming dogs started from rest.

$v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s$

and the displacement is given by:

$d=\frac{1}{2}*a*t^2\\d=7.3m$

Using the energy conservation formula:

$K_i+U_i+W_d+W_f=K_f+U_f$

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

$W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J$

so:

$W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J$

The power is given by:

$P=\frac{W}{t}\\\\P=\frac{2457J}{4.4s}\\\\P=558W$

and the force exerted by the dogs:

$W_d=F_d*d*cos(\theta)\\F_d=\frac{W_d}{d*cos(0)}\\\\F=\frac{2457J}{7.3m*(1)}\\\\F=337N$

4 0

3 years ago

Please help me with this 29 points

Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

7 0

2 years ago

What happens when an electron moves from an excited state to the ground state?

Darina [25.2K]

<span>When an electron moves from an excited state to the ground state, "Energy releases"

Hope this helps!</span>

6 0

3 years ago

un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia

Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

$R=\frac{\rho\times l}{A}$

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

$R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm$

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

4 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

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