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Ostrovityanka [42]

2 years ago

12

A rock has 750 J of potential energy as it sits on a ledge. If the rock were pushed off the ledge, how much would it have just b

efore it hit the ground?

1 answer:

wariber [46]2 years ago

8 0

Answer:

the 750 j will have potential energy is 375

Explanation:

750/2 is 375

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

vlada-n [284]

Answer:

Explanation:

The process is isothermic, as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv = m / M RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0

3 years ago

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc

guajiro [1.7K]

Answer:

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

$a_o=a\times e^{-kt}$

where,

k = rate constant

t = age of sample

$a_o$ = let initial amount of the reactant

a = amount left after decay process

We have :

$a_o=x$

$a=58\%\times x=0.58x$

t = 95 s

$0.58x=x\times e^{-k\times 95 s}$

$\k= 0.005734 s^{-1}$

Half life is given by for first order kinetics::

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{0.005734 s^{-1}}=120.86 s$

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0

2 years ago

I NEED HELP WITH THIS QUESTION ASAP PLEASE!

daser333 [38]

Wood isn’t a medium. Pls give brainliest

8 0

2 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

The even distribution of weight in a compostion is known as

mojhsa [17]

The answer is Balance. hopei helped? ahahahaha

3 0

3 years ago