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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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A 150 kg car hits a wall with 45 N of force, what was its acceleration?​
alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: a = \frac{F}{m}

Solution: To find a, divide the force by the weight  

A = F ÷ m

=  45 ÷  150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

4 0
3 years ago
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What
FromTheMoon [43]

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

7 0
3 years ago
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