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stellarik [79]
3 years ago
15

3. A woman drove her car from home to her daughter's school. The odometre on her dashboard says she travelled 4.5 km to do this.

She then immediately drove back home, using a different route, which was 5.5 km long. The whole journey took 30 minutes.
a. What distance did she travel?

b. What was her displacement?

C. What was her average speed during the journey?
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

Look Below -->

Explanation:

a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )

b. Her displacement is  0 km because she went back home. (Displacement is the difference between the end and starting points)

c. 3 km/hr (30 minutes / 10 km)

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.
SpyIntel [72]

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Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

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So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by \tau =F\times d=45.5\times 0.0245=1.11475N-m

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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
3 years ago
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