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3 years ago

1. Calculate the momentum of a 1,500 kg car traveling at 6 m/s.

Chemistry

1 answer:

ziro4ka [17]3 years ago

8 0

Answer:

1. $9000kgm/s$

2. $258,000kgm/s$

Explanation:

1.Momentum is given as the product of mass by velocity of an object.

Momentum, $p=mv$

m=1,500kh, v=6m/s

$p=1500*6\\p=9000kgm/s$

2.Momentum, $p=mv$

m=7800kg, v=30m/s

$p_1=7800*30\\p_1=234,000kgm/s\\$

new mass=7800+800=8600

As mass is increased, so does the resultant velocity as mass is directly proportional to velocity.

$p_2=8600*30\\p_2=258,000kgm/s$

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Please help!!

hichkok12 [17]

It’s 65% average that your answer

6 0

2 years ago

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

max2010maxim [7]

Answer:

The answer to your question is: Initial temperature of copper = 67.1°C

Explanation:

Data

mass Copper = 248 g

volume Water = 390 ml

T1 water = 22.6°C

T2 = 39.9°C

T1 copper = ?

Specific heat water = 1 cal/g°C

Specific heat copper = 0.092 cal/g°C

Formula copper water

Heat is negative for copper because it releases heat

- mCp(T2 - T1) = mCp(T2 - T1)

- (248)(39.9 - T1) = 390 (1)((39.9 - 22.6) Substitution

-9895.2 + 248T1 = 390(17.3) Simplification

-9895.2 + 248T1 = 6747

248 T1 = 6747 + 9895.2

248 T1 = 16642.2

T1 = 16642.2 / 248

T1 = 67.1 °C Result

6 0

3 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago

Based on the diagram below, how much of the excess reactant is left over? *

pantera1 [17]

Answer:

3 pieces of lunch-meat and 2 slices of cheese

Explanation:

You have enough bread to make 3 sandwiches

You have enough lunch-meat to make 4 sandwiches

You have enough cheese to make 5 sandwiches

In all you have enough material to make 3 sandwiches

so if you subtract three from each number above you will have no bread, enough lunch-meat to make one sandwich and enough cheese to make two sandwiches

luch-meat for one sandwich is: 3 pieces

Cheese for two sandwiches is: 2 pieces

5 0

3 years ago

2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.

Savatey [412]

<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

☆...hope this helps...☆

_♡_mashi_♡_

5 0

1 year ago