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myrzilka [38]
3 years ago
11

What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Physics
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N

The net power needed to change the speed of the vehicle is calculated as;

P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W

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A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Anastaziya [24]

Answer:

The gauge pressure is  P_g  =  2058 \ P_a

Explanation:

From the question we are told that

       The height of the water contained is  h_w  =  30 \ cm  =  0.3 \ m

        The height of liquid in the cylinder is  h_t  =  40 \ cm  = 0.4 \ m

       

At the bottom of the cylinder the gauge pressure is  mathematically represented as

        P_g  =  P_w + P_o

Where  P_w is the pressure of water which is mathematically represented as

      P_w  =  \rho_w  *  g * h_w

Now  \rho_w is the density of water with a constant values of  \rho_w  = 1000 \ kg /m^3

   substituting values

      P_w  = 1000 *  9.8 *  0.3

     P_w  =  2940 \  Pa

While P_o is the pressure of oil which is mathematically represented as

          P_o  =  \rho_o *  g *  (h_t -h_w )

Where \rho _o is the density of oil with a constant value

         \rho _o  = 900 \ kg / m^3

substituting values

       P_o  =  900 *  9.8 * (0.4 - 0.3)

       P_o  =  882 \ Pa

Therefore

      P_g  =  2940 - 882

      P_g  =  2058 \ P_a

6 0
3 years ago
Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We
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Answer:

q=\frac{mg}{E_o}              

Explanation:

Given:

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The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

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This is balanced by the weight, mg

Equating the two:

qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}

4 0
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