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2 years ago

What is dimension of volume flow rate.?

Physics

1 answer:

ella [17]2 years ago

6 0

<span> is the volume of fluid which passes per unit time; usually represented by the symbol Q (sometimes V̇).</span>

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A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i

scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0

3 years ago

A mover pushes a 30 kg crate across a level florr with a force of 250 N, but the crate accelerates at a rate of only 5.83 m/s sq

arlik [135]

Force of friction=75 N

Explanation:

we use Newton's second law of motion

F- Ff= ma

F = applied force=250 N

Ff= force of friction

a= acceleration= 5.83 m/s²

m= mass=30 kg

250- Ff= 30 (5.83)

Ff=250-30(5.83)

Ff=250-175

Ff=75 N

7 0

2 years ago

How many hydrogen atoms are in the following formula:

True [87]

C it have two atoms for hydrogen

6 0

1 year ago

27. The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. Please use 2 conten

KIM [24]

Answer:

Once we place a positive test at a point close to the sphere, we find that an electrostatic force is applied to the outside of the sphere. Therefore, at any point around the sphere, the electric field vector is radially outward.

4 0

2 years ago

A projectile (mass = 0.15 kg) is fired at and embeds itself in a stationary target (mass = 2.44 kg). With what percentage of the

Bess [88]

Answer:

38.6 %

Explanation:

First of all, we have to calculate the final velocity of the block-bullet system. We can apply the law of conservation of momentum:

$m u + M U = (m+M)v$

where

m = 0.15 kg is the mass of the bullet

u is the initial velocity of the bullet

M = 2.44 kg is the mass of the block

U = 0 is the initial velocity of the block (it is at rest)

v is the final velocity of the bullet+block

Solving for v,

$v=\frac{mu}{m+M}$

The total initial kinetic energy of the system is just the kinetic energy of the bullet:

$K = \frac{1}{2}mu^2$

While the final kinetic energy of the block+bullet is:

$K' = \frac{1}{2}(m+M) v^2 = \frac{1}{2}(m+M) \frac{(mu)^2}{(m+M)^2}=\frac{1}{2} \frac{m}{m+M}u^2$

So the fraction of kinetic energy lost is

$\frac{K-K'}{K}=\frac{\frac{1}{2}mu^2 - \frac{1}{2}\frac{m}{m+M}u^2}{\frac{1}{2}mu^2}=\frac{m-\frac{m}{m+M}}{m}=\frac{0.15-\frac{0.15}{0.15+2.44}}{0.15}=0.614$

And so, the fraction of kinetic energy left in the projectile after he flies off the block is

1 - 0.614 = 0.386 = 38.6 %

4 0

2 years ago