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S_A_V
1 month ago
8

In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the

fulcrum.
Physics
1 answer:
serg [3.5K]1 month ago
4 0
The full sentence states:
In a third class lever, the distance between the effort and the fulcrum is LESS than the distance between the load/resistance and the fulcrum.
In a third class lever, the fulcrum is positioned on one end of the effort, while the load/resistance is on the opposite side, placing the effort somewhere in between. Consequently, the distance from the effort to the fulcrum is less than that from the load to the fulcrum.
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what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
Sav [3153]
Recall this formula for a device operating in a direct current circuit:
P = IV
In this equation, P stands for the power emitted by the device, I signifies the current passing through the device, and V represents the voltage drop across it.

Using ampere for current and volt for voltage means that multiplying current by voltage gives you power measured in watts.
5 0
2 months ago
At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w
serg [3582]

Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

               E = \frac{\sigma}{2 \epsilon_{o}}

where,   \sigma is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

             F = eE

where,    e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

                       F = ma

                 a = \frac{eE}{m}

                    = \frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}

                     = \frac{e \sigma}{2m \epsilon_{o}}

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

              v_{f} = v_{i} + at

                     = (0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t

                     = \frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}

                     = 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

                v' = \sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}

                    = \sqrt{921600 + 467820.43}

                    = \sqrt{1389420.43}

                    = 1178.73 m/s

Consequently, we conclude that the proton's speed is 1178.73 m/s.

3 0
1 month ago
Un tren parte de la ciudad A, a las 8 h. con una velocidad de 50 km/h, para llegar a la ciudad B a las 10 h. Allí permanece dura
Keith_Richards [3271]

Response:

AB = 100 km; BC = 80 km; AC = 180 km

Time of arrival = 11:30

Reasoning:

1. Distance from A to B

(a) Duration of travel

Duration = 10:00 - 8:00 = 2.00 hours

(b) Distance

Distance = speed × time = 50 km/h × 2.00 h = 100 km

2. Distance from B to C

Distance = 80 km/h × 1 h = 80 km

3. Summary of Distances

AB = 100 km

BC = 80 km

AC = 180 km

4. Time of Arrival

Departure from A = 08:00

Travel duration to B = 2:00

Arrival at B = 10:00

Waiting time at B = 0:30

Departure from B = 10:30

Travel duration to C = 1:00

Arrival at C = 11:30

8 0
2 months ago
On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
ValentinkaMS [3465]

Answer:

The equivalent distance in kilometers is 4012 ×10^{-6} km.

Explanation:

It's known that 1 millimeter converts to 10^{-3} meters. Then, 1 meter converts to 10^{-3} kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = 10^{-3} m

1 m = 10^{-3} km

Thus, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 ×10^{-6} km.

The distance therefore is 4012 ×10^{-6} km.

5 0
2 months ago
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Keith_Richards [3271]
25.82 m/s Explanation: Given: Force applied by the baseball player; F = 100 N Distance the ball travels; d = 0.5 m Mass of the ball; m = 0.15 kg To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved. It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows: F × d = ½mv² 100 × 0.5 = ½ × 0.15 × v² Solving gives: v² = (2 × 100 × 0.5) / 0.15 v² = 666.67 v = √666.67 v = 25.82 m/s.
4 0
1 month ago
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