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faltersainse [42]
3 years ago
8

Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Y: attractive and repulsive Z: very small range X: attractive only Y: very small range Z: attractive and repulsive X: attractive and repulsive Y: infinite range Z: attractive only X: very small range Y: attractive only Z: infinite range
Physics
1 answer:
Maslowich3 years ago
8 0

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive  or attractive only

Y: Very small range

Z:  Repulsive and attractive

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A baseball player hits a 140 g baseball with a force of 2800 N. What is the
Murljashka [212]
B because 2800 divide by 40 is 20
6 0
2 years ago
Two movers are pushing a large crate with a force of 60.0 n each. one pushes north, the other east. what is the equilibrant forc
romanna [79]
To get the solution you must need to draw a force triangle. Attach the head of the 60N north force arrow with the tail of the 60N east force arrow. The subsequent is the arrow connecting he tail and head of the two arrows. 
You get a right angled triangle, and the resultant is (60^2 + 60^2) ^0.5 = 84.85 N or 85 N northeast.
8 0
3 years ago
If there is a huge boulder on your lawn and you want to determine its density, but it is
Olin [163]

There are different options here but all of them work by approximating and assuming.

i) that the boulder is above ground.

ii) that the bottom surface of the boulder is known.

iii) the shape of the boulder is taken into account.

The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.

All the above methods are estimating methods.

*Another way to find the density is through specific gravity.

S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>

Density of water

If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.

If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.

This is what I think after correction and allthe best!

3 0
2 years ago
Read 2 more answers
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0
3 years ago
When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their
AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

F = qvB

Now this causes a centripetal acceleration

F =  = mv^2/r

 so

qvB = mv^2/r ...........(i)

B = mv/(rq)  ...............(ii)

If  accelerating potential V is  same and  then  kinetic energy equals the potential energy difference

\frac{1}{2} mv^2 = Vq

v = \sqrt{(2Vq/m)}      put these value in equation (ii)

B = m\frac{\sqrt{(2Vq/m)} }{rq}  

simplifying we get  

B =m \frac{(\sqrt{ 2Vm/q})}{r}

for same location r will be same in both case

B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}      ..............(iii)

B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}    ..........(iv)

 dividing (iv) and (iii) equation we get

\frac{B_{10}}{B_{7}}   =   \sqrt{\frac{m_{10}}{{m_7}} }

{B_{10}}  =  B_{7}  \sqrt{\frac{m_{10}}{{m_7}} }

B_{10}       = 0.2574T\sqrt{\frac{  (1.663x10^-26}{(1.165x10^-26)}

so on solving we get  

             =0.3075 T

5 0
3 years ago
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