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2 years ago

What is another way to describe the vector below?

Physics

2 answers:

nignag [31]2 years ago

8 0

Answer:

-40 feet to the left

Explanation:

opposite of 40 feet from right is -40 to left

IrinaVladis [17]2 years ago

8 0

Answer:

B. -40 feet to the left

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A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passin

Vera_Pavlovna [14]

Answer:

V= 3.48 m/s

Explanation:

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2 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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2 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

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2 years ago

What would be the best design for an experiment that tests how much water expands when frozen?

galben [10]

B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

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1 year ago

two tugboats pull a disabled supertanker. each tug exerts a constant force of 2.20×106 n , one at an angle 19.0 ∘ west of north,

asambeis [7]

The amount of work the two tugboats completed on the supertanker W = 3.12 × 10^9 joules .

Work is a physics term used to describe the energy transfer that takes place when an object is moved over a distance by an external force, at least some . The component of the force acting along the path multiplied by the length of the path can be used to calculate work if the force is constant. Mathematically, this idea is expressed as W = fd, where W is the work and f is the force multiplied by d, the distance. Work is completed when the force is applied at an angle of with respect to the displacement. Performing work on a body involves moving it in its entirety from one location to another as well as other methods. However, the work is thought to be negative if the applied force is in opposition to the item's motion, indicating that energy is being pulled away from the object

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brainly.com/question/1374468

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8 months ago