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BigorU [14]
3 years ago
13

Compare the signs of ƒ for lenses and mirrors.

Physics
2 answers:
ELEN [110]3 years ago
4 0

Answer:

A positive focal point for concave mirrors and convex lenses.

A negative focal point for convex mirrors and concave lenses.

Explanation:

STALIN [3.7K]3 years ago
3 0

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

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a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.
inessss [21]

Answer:

 B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

           B = - \frac{P}{ \frac{\Delta V}{V} }

           

The negative fate is for the module to be positive since the volume change is negative

       

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

             B = - \frac{10^4}{ \frac{-0.20}{5000} } = \frac{10^4}{4 \ 10^{-5} }

             B = 2.5 10⁸ Pa

4 0
3 years ago
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
3 years ago
A box weighing 200N is pushed on a horizontal floor. What acceleration will result if a horizontal force of 100N is applied on t
Bogdan [553]

Answer:

the friction force in the reverse direction is 200 *0.4=80 N.

the net forward force acting on the box is therefore

Fnet= 100 - 80  N

= 20 N

acceleration = Fnet / mass

=Fnet *g/(weight)

=20 *9.8/200 = 0.98 m/s^2

Explanation:

4 0
3 years ago
Read 2 more answers
IM SOO CONFUSED PLS HELP!! The mass of the nucleus is approximately EQUAL to the mass number multiplied by ____ Atomic Mass unit
nevsk [136]

Answer:

option a.

Explanation:

We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.

So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.

Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u

Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u

(where u = atomic mass unit)

Then the weight of the nucleus is about A times 1u, or:

A*1u = A atomic mass units.

Then the correct option is:

The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.

option a.

5 0
3 years ago
The most common atom used in fission is ​
Eduardwww [97]

Answer:

Uranium

Explanation:

5 0
3 years ago
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