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swat32
1 month ago
14

Use this free body diagram to help you find the magnitude of the force F2 needed to keep this block in static equilibrium. WILL

GIVE BRAINLIEST

Physics
1 answer:
Yuliya22 [3.3K]1 month ago
4 0
In static equilibrium, all forces balance out. Therefore, to simplify, start by breaking down F1 into its horizontal and vertical components. Since no other forces act horizontally, F1's horizontal component is known to be 40N. This information can be used to determine the vertical component using the Pythagorean theorem. Once the components are established, simply add the vertical components to calculate the difference between the upward and downward forces.

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Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th
Softa [3030]

Response:

The new resistance is half of the original resistance.

Explanation:

Resistance in a wire is represented by:

R=\dfrac{\rho L}{A}

\rho = resistivity of the material

L and A are the physical dimensions

If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r

The updated resistance of the wire can be calculated as follows:

R'=\dfrac{\rho L'}{A'}

R'=\dfrac{\rho (2L)}{\pi (2r)^2}

R'=\dfrac{1}{2}\dfrac{\rho L}{A}

R'=\dfrac{1}{2}R

The new resistance equals half of the original resistance. Thus, this provides the solution needed.

4 0
2 months ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is \sqrt{725} ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t} = \frac{725}{\sqrt{725}} = \sqrt{725} ft / sec

5 0
2 months ago
A paper in the journal Current Biology tells of some jellyfish-like animals that attack their prey by launching stinging cells i
Yuliya22 [3333]
Given that, the starting speed of the cells is 0 since they were at rest. The cell's acceleration is specified, along with time t = 700 ns. We aim to calculate the peak speed achieved by the cells and the distance covered during the acceleration. Let v signify the final velocity. Let d represent the distance traversed. We'll apply the equations of motion to find the solution.
3 0
2 months ago
The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p
Maru [3345]
True. Explanation: In this instance, the area of the graph represents the impulse. Impulse is defined as the change in an object's momentum. Moreover, it is also expressed as the product of the force acting on an object and the duration of the impact. When we graph the force against time, if the force remains constant, the resultant graph will take on a rectangular shape, and the area under that graph will equal the impulse's definition.
8 0
2 months ago
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