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3 years ago

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Please help on this one

1 answer:

ale4655 [162]3 years ago

4 0

That thing is used as a lever.

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What is the role of electrical forces in nuclear fission

77julia77 [94]

The electrical forces pulls nucleus apart

8 0

3 years ago

two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

6 0

2 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

Substituting values we have

$\Delta V = 0.510 -0.160$

$= 0.350L$

Generally workdone is mathematically represented as

$W = -P \Delta V$

W is negative because the working is done on the environment by the system which is indicated by volume increase

Substituting values

$W = - 5* 0.350$

$= -1.75 \ L \ \cdot atm$

Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

$= -177.275J$

7 0

2 years ago

Read 2 more answers

Can someone help me label this flower??

tensa zangetsu [6.8K]

Hope this helps!!!!!!!!!!!!!

7 0

2 years ago