All categories

3 years ago

A negative velocity, approaching zero, represents a negative acceleration. True or False

Physics

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

False.

Explanation:

Lets assume our positive direction to the right (this reasoning works for any direction). A negative velocity would then be then directed to the left. If it varies as such that it aproaches to zero, it means that the variation is directed to the right, and that is where the direction of the acceleration must be pointing. In other words, its losing its velocity, so the acceleration must point opposite to the velocity. Then it means the acceleration is positive.

You might be interested in

How much energy is required to move an electron through a potential difference of

Tresset [83]

I think it’s going to be the 2nd one

3 0

3 years ago

How are the surface of Mercury and Mars similar to the surface of the Earth?

scZoUnD [109]

Answer:

The terrestrial planets all have rocky surfaces that feature mountains, plains, valleys and other formations. ... Mars has very low atmospheric pressure, and Mercury has almost none, so craters are more common on these planets.

Explanation:

8 0

3 years ago

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for

Monica [59]

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

3 0

3 years ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$