When the temperature of a certain solid, rectangular object increases by AT, the length of one

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?

Romashka [77]

When you stand at the center of curvature in front of a concave mirror, the image created will be formed at the center of curvature also, but it will be inverted. The image formed will be real, and will have the same size as you.

3 0

3 years ago

The beat your doctor listens to through a sethoscope is the sound of the four values opening

djyliett [7]

and closing .

The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.

The mitral valve is located between the left atrium and the left ventricle. It closes the left atrium to collect oxygenated blood from the lungs and opens to pass it on to the left ventricle.

The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.

The aortic valve is located between the aorta and the left ventricle. It closes the left ventricle and opens to the aorta to pass on the oxygen-rich blood to the body.

The pulmonary valve is located between the pulmonary artery and the right ventricle. It closes off the right ventricle and opens to pass on unoxygenated blood to the lungs.

5 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

4 years ago

Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of

KonstantinChe [14]

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

$m_{s}c_{w} \alpha _{w}$ + $m_{s}l$ = $m_{w}c_{w} \alpha _{w}$ + $m_{c}c_{c} \alpha c_{c}$

L=specific latent heat of water(steam)=2268J/g

$c_{w}$ =specific heat capacity=4.2J/gK

$c_{c}$ =specific heat capacity of calorimeter =0.9J/gk

$m_{w}$ =280g

$m_{c}$ =38g

α=change in temperature

$\alpha _{c}$ =(40-25)=15

$\alpha _{w}$ =(40-25)=15

$\alpha _{s}$ =(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make $m_{s}$ the subject of formula

$m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}$

Hence

$m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g$

Hence the amount of steam needed is 7.2g

6 0

3 years ago

What change will always result in an increase in gravitational force between two objects

tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

$F = G\frac{m_1m_2}{d^2}$

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0

4 years ago