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LenKa [72]
3 years ago
6

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant

speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
Physics
1 answer:
levacccp [35]3 years ago
3 0

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V_B = 54 km/hr

V_A = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω_B = V_B / α

so, we substitute

ω_B = ( 54 × 1000/3600) / 100

ω_B = 15 / 100

ω_B = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

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Here, t = 2 s  [ Given ]
a = 9.8 m/s²  [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

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4 0
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. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of
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The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

F_f = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

W=580 N (weight of the leaves)

Substituting, we find:

F=(0.55)(580)=319 N

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I think it is the Milky Way.
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If τ=r×F then F.τ is equal
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A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres
Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

                               V = L^3  = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

                               P*V = n*R*T

                                n = P*V / R*T

                                n = 101.3*10^3 * 0.001 / 8.314*300

                                n = 0.04061 moles

- Compute the mass of the gas:

                                m = n*M

                                m = 0.04061*28.9

                                m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

                                F_g = m*g

                                F_g = 1.174*9.81*10^-3

                               F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

                                F_c = P*A

                                F_c = (101.3*10^3)*(0.1)^2

                                F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

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