The amount of light that enters the pupil is controlled by the:<br> retina.<br> lens.<br> inis.

A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

lbvjy [14]

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

3 0

2 years ago

Bowling ball and a balloon are dropped from a 50 m tall building at the same time. IN AIR which object will hit the ground firs

HACTEHA [7]

It depends, What is the balloon filled with? Just air, heilum, water, etc. If its just air then the bowling ball would most likly hit first, but if the balloon is filled with water it would problably hit at the same time, theres also air resistace to think about is there any or it is almost like a vaccum chamber?

8 0

2 years ago

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
$F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N$

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

6 0

3 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago

Please help me I have to send for my teacher

expeople1 [14]

Answer:

for the fill in the blanks

1- static

2-kinetic

3-coeffiecient

4-opposite to

5-sin theta

6-cos theta

im not sure however what to do with the top part or if its even part of what you need help with

8 0

2 years ago

The amount of light that enters the pupil is controlled by the: retina. lens. inis.