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Naddik [55]
2 years ago
9

A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the ab

sence of friction. At what angle should the curve be banked? 2. what is the correct equation for the sum of the forces in the x direction? What is the correct equation for the sum of the forces in the y direction? A. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ−mg=0 B. ΣFx:−Ncosθ=−mv2/R; ΣFy:Nsinθ−mg=0 C. ΣFx:Nsinθ=mv2/R; ΣFy:Ncosθ+mg=0 D. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ=0 E. ΣFx:−Nsinθ=−mv2/R; ΣFy:Ncosθ−mg=0
Physics
1 answer:
belka [17]2 years ago
6 0

Answer:

A)   θ = 13.1º  , B)  E

Explanation:

A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.

In this reference system the only force that we must decompose is the Normal one, let's use trigonometry

        sin θ = Nₓ / N

        cos θ = N_{y} / N

        Nₓ = N sin θ

       Ny = N cos θ

x-axis (radial)

        Nₓ = m a

where the acceleration is centripetal

         a = v² / R

we substitute

        -N sin θ = -m v² / R                   (1)

the negative sign indicates that the force and acceleration towards the center of the circle

y-axis (Vertical)

          Ny - W = 0

           N cos θ = mg

           N = mg / cos θ

we substitute in 1

          mg / cos θ  sin θ = m v² / R

          g tan θ = v² / R

          θ = tan⁻¹ (v² / gR)

we calculate

        θ = tan⁻¹ (25² / 9.8 274)

        θ = 13.1º

B) when comparing the equations the correct one is E

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Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-
Nady [450]

Answer:

a)   W_total = 8240 J , b)  W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

       W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

       W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

          F - W = m a

          W = mg

          F = m (a + g)

          F = 80 (1 + 9.8)

          F = 864 N

The work of this force we will call it W1

   We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

        F₂ - W = 0

        F₂ = W

        F₂ = 80 9.8

        F₂ = 784 N

The work of this fura we will call them W2

The total work is

         W_total = W₁ + W₂

         W_total = (F + F₂) y

         W_total = (864 + 784) 5

         W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

        W₁ / W₂ = F y / F₂ y

        W₁ / W₂ = 864/784

        W₁ / W₂ = 1.1

7 0
2 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
2 years ago
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initi
Tanzania [10]

Answer:

(C) T

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence the tension.

Note: The figure of question is added in the attachment

3 0
2 years ago
Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?
morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays,  parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

3 0
2 years ago
A 20.0 kg curling stone travels 30.0 m along the ice surface. If the frictional force is 10.0 N, the thermal energy produced is
evablogger [386]
The thermal energy is where the work of friction comes from.  That is what stops it eventually.  In this case a counter force of 10N is applied over the distance of 30.0m.  The energy is given by Force*Distance.  Here this is 300J.  This friction work is the thermal energy.
4 0
2 years ago
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