HELP URGENT———Why is creativity important in constructing scientific methods?

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G

Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time

So

acceleration(-g)= dv/dt

Solve it

dv = a dt

dv = -g dt

v - v₀ = -gt

v= dy/dt

dy = v dt

dy = ( v₀ - gt ) dt

y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0

3 years ago

What is the difference between distance and displacement? Give an example of a situation where distance and displacement both ha

Oksana_A [137]

Explanation :

Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.

Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.

When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.

When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.

4 0

3 years ago

1.When you give one set of washers a downward push, does it move as easily as the other set? Does it stop before it reaches the

Vlad1618 [11]

Answer:No, it doesn't move easily downward because it will try to resist the movement ,due to a resistance force of inertia that it possess at rest.

Explanation:when an object has higher or larger mass it tends to resist any motion given to it unlike the one with lower mass.

The larger the mass the more resistance force an object has.

5 0

3 years ago

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Ainat [17]

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

$\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg$