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il63 [147K]
3 years ago
9

A jogger runs at an average speed of 4.20 mi/h. (a) how fast is she running in m/s? (report your answer to the correct number of

significant figures.) 1.88 m/s (b) how many kilometers does she run in 33.0 min? (report your answer to the correct number of significant figures.) 3.72 km (c) if she starts a run at 11:15
a.m., what time is it after she covers 4.84 × 104 ft? : p.m.
Physics
1 answer:
PolarNik [594]3 years ago
5 0
a) How fast is she running in m / s?
 The first thing you should know for this case is the following conversion
 1mi = 1609.34m
 1h = 3600s
 Applying the conversion
 4.20 (mi / h) * (1609.34 / 1) (m / mi) * (1/3600) (h / s) = 1.88 m / s
 answer
 She is running 1.88 in m / s

 (b) how many kilometers does she run in 33.0 min? 
 By definition, the distance equals the speed by time
 d = v * t
 Also,
 1min = 60seconds
 1 km = 1000 meters.
 We have then
 d = (1.88) * (33) * (60) * (1/1000) = 3.72 Km
 answer
 she run 3.72 Km in 33.0 min
 
(c) if she starts a run at 11:15 a.m., what time is it after she covers 4.84 × 104 ft? : p.m.
 The first thing we need to know is the following conversion
 1Km = 3280.84 feet
 We have then
 3.72 (Km) * (3280.84 / 1) (feet / Km) = 12204.73 feets
 I have the following rule of three we can know how long she travels 4.84 × 104 ft
 12204.73 feets ----> 33 min
 4.84 * 10 ^ 4 feets ----> x
 Clearing x:
 x = ((4.84 * 10 ^ 4) / (12204.73)) * (33) = 131 min
 131 min = 2:11:00 hours
 she starts a run at 11:15 a.m.
 she finishes at
 1:26 p.m.
 answer
  it is 1:26 p.m when she covers 4.84 × 104 ft
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Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

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3 years ago
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Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

h = \frac{gt^2}{2}

solving for t

we have

t = \sqrt{\frac{2h}{g}}

substituing all value for time t

t = \sqrt{\frac{2\times 11.98}{9.8}}

t = 1.56 s

we know that speed is given as

v = \frac{d}{t}

v =\frac{11.98}{1.56}

v = 7.67 m/s

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