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solong [7]
3 years ago
6

Which type of particle is the basic building block of matter?

Physics
2 answers:
Basile [38]3 years ago
8 0

Atom is the basic building block of matter .

Elena L [17]3 years ago
7 0

Answer:

B

Explanation:

The Atom

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A common magnifying glass is an example of.​
Misha Larkins [42]

Answer:

planoconvex lens

Explanation:

8 0
3 years ago
Read 2 more answers
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Goryan [66]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

4 0
3 years ago
A 0.23-f capacitor is desired. What area must the plates have if they are to be separated by a 3.8-mm air gap?
tia_tia [17]

The area of the plates must have is(A)= 9.91×10⁷ m²

<h3 /><h3>How can we calculate the value of a area of a capacitor?</h3>

To calculate the the value of a area of the plates of a capacitor, we are using the formula,

C=\frac{\epsilon_0 A}{d}

Or, A= \frac{C\times d}{\epsilon_0}

Here we are given,

C= The desired capacitance of a capacitor.

= 0.23F

d=distance of separation between the plates.

=3.8mm= 0.0038m.

\epsilon_0= permittivity of the vacuum.  

=8.854×10⁻¹²F/m

We have to calculate the area of the plates must have = A m².

Now we put the known values in the above equation, we can get

A= \frac{C\times d}{\epsilon_0}

Or, A=\frac{0.23\times 0.0038}{8.854\times 10^{-12}}

Or, A= 9.91×10⁷ m²

From the above calculation, we can conclude that the area of the plates must have is(A)= 9.91×10⁷m²

Learn more about Capacitor:

brainly.com/question/13578522

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5 0
2 years ago
A cold coke bottle is on the pan of a balance and is left open. What happens to its weight?​
Arlecino [84]

Answer:

It's weight would decrease over time.

Explanation:

The cold coke is going to stabilise in the room temperature. This causes the coke to heat up to room temperature. When the coke heats up, some molecules acquire enough kinetic energy that the evaporate. The evaporation of the molecules causes a decrease in weight.

Hope it helped!

3 0
3 years ago
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A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
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