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Alecsey [184]
2 years ago
5

How do u find the number of electronic

Physics
1 answer:
vichka [17]2 years ago
6 0
Count.................
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A force of 150N at an angle of 60 degree to the horizontal to pull a box through a distance of 50m calculate the work done
Alexxandr [17]
  • Force=150N
  • Angle=60°
  • Displacement=50m

\boxed{\sf W=Fscos\Theta}

\\ \sf\longmapsto W=150(50)cos 60

\\ \sf\longmapsto W=7500\times \dfrac{1}{2}

\\ \sf\longmapsto W=3750J

6 0
3 years ago
Describe Kinetic Energy.​
mina [271]
In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

4 0
3 years ago
A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0
2 years ago
A truck driving east on a highway goes from the 42 mi marker to the 48 mi
Leona [35]

Average velocity is 1..2 mi/min east

Explanation:

  • Velocity = Displacement/Time

Here, displacement = 48 mi - 42 mi = 6 miles

Time = 5 minutes

⇒ Average Velocity = 6/5 = 1.2 mi/min east

3 0
3 years ago
Create a model in the space below that demonstrates how action potentials ensue. A sample model can be found under the Unit Pack
Pepsi [2]

Answer:

K+NA+30

Explanation:

8 0
3 years ago
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