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nlexa [21]
3 years ago
14

g An electric hot plate raises its own internal energy and the internal energy of a cup of water by 9000J, and there is at the s

ame time 2700J transferred to the cooler air. How much energy was transferred to the hot plate in the form of electricity
Physics
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

11700j

Explanation:

add the two because the plate has to maintain the temp.

2700+9000=11700

You might be interested in
lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu
Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

Therefore, the total work done is 578.123 J and the answer is option E

6 0
1 year ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?
Serhud [2]

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

5 0
3 years ago
A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and
ycow [4]

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

4 0
3 years ago
Explanation for x=73km/h•0.5s= 10.14 pls help how did they get 10.14? What was the work around it
Sergio [31]
Hi I am the guy to get mad at
4 0
3 years ago
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