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klio [65]
3 years ago
12

A particle is moving with a velocity of 60.0 m/s in the positive x direction at t= 0. Between t= 0 and t= 15.0 s the velocity de

creases uniformly to zero. What was the acceleration during this 15.0-s interval? What is the significance of the sign of your answer?
Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

Acceleration of the object is given as

a = - 4 m/s^2

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

Explanation:

As we know that velocity decreases uniformly from maximum to zero

so here in this uniform deceleration we know that

v_i = 60 m/s

v_f = 0

t = 15 s

now by using the equation of kinematics we have

v_f - v_i = at

0 - 60 = a(15)

a = - 4 m/s^2

so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
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Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

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A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
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Answer:

option C

Explanation:

given,

force act on west  = 20 lb

force act at 45° east of north = 80 lb

magnitude of force = ?

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magnitude of forces in x- direction

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F = \sqrt{F_x^2+F_y^2}

F = \sqrt{56.57^2+36.57^2}

F = 67.36 lb≅ 67 lb

hence, the correct answer is option C

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