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Novosadov [1.4K]
3 years ago
6

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 15.0 min at 50.0 km/

h, 19.0 min at 85.0 km/h, and 55.0 min at 60.0 km/h and spends 25.0 min eating lunch and buying gas.
a. What is the total distance traveled over the entire trip (in km)?
b. What is the average speed for the entire trip (in km/h)?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer

given

Case 1-Speed of car 50 Km/h for 15 min

Case 2-Speed of car 85 Km/h for 19 min

Case 3 - Speed of car 60 km/h for 55 min

lunch break time = 25 min

a) total distance travel

   we know,

    distance = speed x time

 D = 50 \times \dfrac{15}{60} + 85 \times \dfrac{19}{60} + 60\times \dfrac{55}{60}

  D = 94.42 Km

b) average\ speed = \dfrac{total\ distance}{total\ time}

   average\ speed = \dfrac{94.42}{\dfrac{15}{60}+\dfrac{19}{60}+\dfrac{55}{60}+\dfrac{25}{60}}

average\ speed = 49.69\ km/h

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Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

6 0
3 years ago
A car accelerates for 10 seconds. During this time, the angular
bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

final angular velocity, \omega_f = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0
3 years ago
Which of the following statements are true?
inessss [21]

Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

TRUE

Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.

d. If an object's acceleration is zero, its velocity must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant

e. If an object's speed is constant, then its velocity must be constant.

FALSE

Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change

7 0
3 years ago
Henry can lift a 200 N load 20 m up a ladder in 40 s. Ricardo can lift twice the load up one-half the distance in the same amoun
prohojiy [21]
Henry will lift 200 N load 20 m up a ladder in 40 s.  While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance. 
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3 years ago
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Select the correct answer. An airplane is flying at a constant speed in a positive direction. It slows down when it approaches t
Triss [41]

An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going to land. this is an example of negative acceleration (D).

7 0
3 years ago
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