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3 years ago

What effect does food colour have on the amount of food fish eat

2 answers:

8 0

Absolutely none. Fish take food into their mouths and taste it. If they like it, they swallow, if not they will spit it out. The color of the flakes has no bearing on what the fish takes in and tastes in my experience.
I hope this helps : )

Illusion [34]3 years ago

8 0

None, like they said fish take food into their mouth and taste it

You might be interested in

Use the collision theory to explain how increasing the temperature of a reaction will affect the rate of the reaction.

kirill115 [55]

Increasing the temperature causes the particles in the reaction to become kinetically excited, hitting one another in increasing frequency. Increased collision among means faster rate or reaction.

7 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

3 years ago

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0

3 years ago

The circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N. Does the centripetal force do any

Snezhnost [94]

Answer:

Work done is 0.

Explanation:

Given that,

The circumference of an orbit for a toy on a string is 18 m, r = 18 m

Centripetal force, F = 12 N

In the circular path, the centripetal force is always perpendicular to the motion of the object. Thus it makes an angle of 90 degrees with the force and displacement. Hence, we can say that the centripetal force does not do any work on the toy when it follows its orbit for one cycle.

5 0

2 years ago

Read 2 more answers

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0

3 years ago