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Law Incorporation [45]
3 years ago
10

A trip is taken that passes through the following points in order

Physics
1 answer:
AURORKA [14]3 years ago
3 0

Answer:

15? I actually don't know

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A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

w = \sqrt{\frac{k}{m}}

w = \sqrt{\frac{250}{0.5}}

w = 22.4 rad/s

f = frequency

Angular frequency is also given as

w = 2 π f

22.4 = 2 (3.14) f

f  = 3.6 Hz

T = Time period

Time period is given as

T = \frac{1}{f}

T = \frac{1}{3.6}

T = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0
4 years ago
Help on springs,I klai
ra1l [238]
K = ∆F / ∆s = 4N / 0.08m = 50N/m
3 0
3 years ago
A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume
n200080 [17]

Answer:

43994

Explanation:

Hope this helps!

7 0
3 years ago
Which human body systems are MOST directly involved when a person is walking?
AysviL [449]

Answer:

The spinal cord is information central in terms of the nerves involved with walking. The spinal nerves in and at the base of the spinal cord directly affect the walking motion.

Explanation:

6 0
3 years ago
Read 2 more answers
An automobile engine takes in 4000 j of heat and performs 1100 j of mechanical work in each cycle. (a) calculate the engine's ef
Semmy [17]
(a) The efficiency of an engine is defined as the ratio between the work done by the engine and the heat it takes in:
\eta= \frac{W}{Q_{in}}
The engine in this problem does a work of W=1100 J and it takes in Q_{in}=4000 J of heat, therefore its efficiency is
\eta= \frac{1100 J}{4000 J}=0.275 = 27.5 \%

(b) The heat taken by the machine is 4000 J; of this amount of heat, only 1100 J are converted into useful work. This means that the rest of the heat is wasted. Therefore, the wasted heat is the difference between the heat in input and the work done by the engine:
Q_{wasted}=Q_{in}-W=4000 J-1100 J=2900 J
7 0
3 years ago
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