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guapka [62]
2 years ago
9

An object in projectile motion will follow which path?

Physics
1 answer:
Tanzania [10]2 years ago
6 0
It will look like an upside down parabola
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Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.
Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

ma = \text{Mass}\times \text{Acceleration}

ma = kg \cdot \frac{m}{s^2}

At the same time we have that

\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2

\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2

\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}

Therefore there is not have same units and both are not consistent and the correct answer is B.

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3 years ago
What are non-metals?<br>explain briefly​
aliya0001 [1]

Answer:

In chemistry, a nonmetal is a chemical element that is mechanically weak in its most stable form, brittle if solid, and usually gains or shares electrons in chemical reactions. There is no universal agreement on which elements are nonmetals; the numbers generally range from fourteen to twenty-three, depending on the criterion or criteria of interest.

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2 years ago
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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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