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3 years ago

6

The cork in Question 1 travels I foot in moving from its lowest position to its highest position. What is the average velocity o

f the cork during this motion?

1 answer:

tatiyna3 years ago

7 0

Answer:

8ft/s

Explanation:

For computing the average velocity of the cork first we have to determine the frequency which is shown below:

$Frequency = \frac{Time\ taken }{one\ minute}$

$= \frac{240}{60\ seconds}$

= 4 Hz

Now the average velocity of the cork is

$\frac{\lambda}{2} = 1 ft\\\\ \lambda = 2ft\\\\ velocity = f\times \lambda \\\\$

= 8ft/s

We simply applied the above formulas to determine the frequency and the velocity of the cork during this motion and the same is to be considered

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Starting from rest and moving in a straight line, a cheetah

UNO [17]

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

$Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}$

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

$Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2$

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

3 0

3 years ago

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0

3 years ago

What property is a characteristic of a pure substance that can be observed with out changing the substance into something else

WARRIOR [948]

physical property is a characteristics of a pure substance that can be observed without changing the substance into someone else.

4 0

3 years ago

A 2.2 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi

iogann1982 [59]

Answer:

a) $k = 2851.2\,\frac{N}{m}$ , b) $v = 0.54\,\frac{m}{s}$

Explanation:

a) According to the Principle of Energy Conservation, the kinetic energy of the box is transformed into elastic potential energy.

$\frac{1}{2}\cdot (2.2\,kg)\cdot (1.8\,\frac{m}{s} )^{2} = \frac{1}{2}\cdot k \cdot (0.05\,m)^{2}$

The spring constant is:

$k = 2851.2\,\frac{N}{m}$

b) The initial speed needed to compress the spring by 1.5 centimeters is:

$\frac{1}{2}\cdot (2.2\,kg)\cdot v^{2} = \frac{1}{2}\cdot (2851.2\,\frac{N}{m} )\cdot (0.015\,m)^{2}$

$v = 0.54\,\frac{m}{s}$

5 0

4 years ago

A truck with a mass of 1810 kg and moving with a speed of 16.0 m/s rear-ends a 673-kg car stopped at an intersection. The collis

nadezda [96]

Answer:

velocity of truck = 7.32 m/s

velocity of truck = 23.32 m/s

Explanation:

mass of truck, M = 1810 kg

initial velocity of truck, U = 16 m/s

mass of car, m = 673 kg

initial velocity of car, u = 0 m/s

Let the final speed of the car is v and the final speed of truck is V.

Use conservation of momentum

Momentum before collision = Momentum after collision

M x U + m x 0 = M x V + m x v

1810 x 16 + 0 = 1810 V + 673 v

28960 = 1810 V + 673 v .... (1)

As the collision is elastic, so the coefficient of restitution is 1.

By using the formula

$e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}$

where, u1 be the initial velocity of truck, u2 be the initial velocity of the car, v1 be the final velocity of truck and v2 be the final velocity of car.

$1=\frac{V - v}}{0-U}$

V - v = - U

V - v = - 16

v - V = 16

v = 16 + V

Substitute in equation (1), we get

28960 = 1810 V + 673 (16 + V)

28960 = 1810 V + 10768 + 673 V

18192 = 2483 V

V = 7.32 m/s

v = 16 + 7.32 = 23.32 m/s

Thus, the velocity of truck after collision is 7.32 m/s and the velocity of car after collision is 23.32 m/s.

4 0

4 years ago