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2 years ago

Please help types of stimuli the sensory organs receive.

Physics

1 answer:

Alborosie2 years ago

3 0

Answer: Photoreceptor cell, Chemoreceptor, Thermoreceptor, Mechanoreceptor

Hope this helps(:

Explanation:

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Determine the magnitude, fda, acting along member da, due to the applied force f=10.0 iâ20.0 j+27.2 k n.

VashaNatasha [74]

Force vector is given as

$F = 10\hat i + 20\hat j + 27.2\hat k$

now we need to find the magnitude of this force

as we know that

$F_x = 10$

$F_y = 20$

$F_z = 27.2$

so here we will find the net force

$F_{net} = \sqrt{F_x^2 + F_y^2 + F_z^2}$

$F_{net} = \sqrt{10^2 + 20^2 + 27.2^2}$

$F_{net} = 35.2 N$

so it is having magnitude as 35.2 N

8 0

2 years ago

Reeti has a mass of 51.0 kg. The Gravitron, a ride that spins so fast that the floor can be removed without the riders falling,

Serggg [28]

Answer:

L = 2774.4 $kgm^{2}/s$

Explanation:

Using the formula

L = mvr

Where L is angular momentum

m is mass

v is velocity

r is radius

m= 51kg , v=17m/s , r=3.2m

L = 51×17×3.2

L = 2774.4 $kgm^{2}/s$

7 0

2 years ago

Physical science includes the study of

liraira [26]

The study of the inorganic world.

4 0

2 years ago

Read 2 more answers

Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?

abruzzese [7]

Answer:

True

Explanation:

When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.

When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus.

It is true.

7 0

2 years ago

Read 2 more answers

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0

2 years ago