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s2008m [1.1K]
2 years ago
5

Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store h

e finds that the store is closed, he instantly turns around and walks back with a speed of 9.0 km/h. Round your final answers to 2 significant figures.
a. What is the magnitude of average velocity and average speed of Greg over the first 30 mins of his travel?
Physics
2 answers:
VladimirAG [237]2 years ago
8 0

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr.  So it takes him (3 km) / (6 km/hr)  =  1/2 hour to get to the store.

That's 30 minutes.  So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.  

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).

frutty [35]2 years ago
5 0

Answer:

Al comment me over here!

Explanation:

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4 0
2 years ago
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Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, L_o = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

L_1 = 35.0cm =  0.35m

So, the change in spring length when mass hangs is

x =  L_1 - L_o

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx =  W =  4.50

k= \frac{4.50}{x} = \frac{4.50}{0.15}

= 30 N / m

b)  Now for the third spring, stretched the length of spring is

L_2 = 50cm = 0.5m

So, the change in spring length is

x'= L_2 - L_o

= (0.5-0.20)m

=  0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

W'= kx'

= 30(0.3)

= 9.0 N

4 0
2 years ago
|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r
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To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

F = ma

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

a = \frac{F}{m}

a = \frac{11*10^3}{80}

a = 137.5m/s

From the cinematic equations of motion we know that

v_f^2-v_i^2 = 2ax

Where,

v_f =Final velocity

v_i =Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

v_f^2-v_i^2 = 2ax

0-v_i^2 = 2(-g)x

v_i =\sqrt{2gx}

v_i = \sqrt{2*9.8*4.8}

v_i = 9.69m/s

From the equation of motion where acceleration is equal to the velocity in function of time we have

a = \frac{v_i}{t}

t = \frac{v_i}{a}

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t = 0.0705s

Therefore the time required is 0.0705s

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lora16 [44]

Answer:

La longitud del camino recorrido es de 25.9 [m]

Explanation:

Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento

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zmey [24]

Answer:

T=366.23\ N

Explanation:

Given:

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Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}

\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }

T=366.23\ N

4 0
2 years ago
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