A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. The forces acting on the crate are

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Answer 1

Answer: are radioactive waves , hope this helps

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Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 2.80g of carbon dioxide is produced from the reaction of 3.4g of octane and 5.9g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
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Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

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Answer: The limiting reagent is oxygen gas.

Explanation:

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For propane:

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=(30.0g)/(44.1g/mol)=0.680mol$

For oxygen:

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=(75.0g)/(32g/mol)=2.34mol$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = $(1)/(5)* 2.34=0.468mol$ of propane

As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent is oxygen gas.

Topic 1 Homework Homework – Due in 13 hours T2HW Question 16 - Challenge Homework – Unanswered The distance from San Francisco to Los Angeles is approximately 385 miles. You and your friends decide to cycle from San Francisco to Los Angeles. If the distance between the cities is about 385 miles and your doctor tells you that you need to drink 1 L of water for every 1 km that you cycle, how many Lof water will each cyclist need to drink on the journey? Enter your answer as a number using 3 significant figures without units. Do not enter the word "liters" as part of your answer. 1609 m = 1.0 mi Numeric Answer Unanswered

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Answer:

Each cyclist will need to drink 619 l

Explanation:

Hi there!!

First, let´s convert the miles to kilometers:

If 1.0 mi = 1609 m, then 385 mi will be:

385 mi · (1609 m/ 1.0 mi) · (1 km/ 1000 m) = 619 km

Now, if each cyclist need to drink one liter water per kilometer ( I think that´s a lot of water!), for the entire journey each cyclist will need to drink:

619 km · 1 l/km = 619 l

Then, each cyclist will need to drink 619 l.

What is shown by a half-reaction?A. oxidation or reduction of an elementB. neutralization of an ion or moleculeC. decomposition of an ion or moleculeD. none of the above

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Answer

A. oxidation or reduction of an element

Explanation

A half reaction can be either oxidation or reduction reaction from a REDOX reaction.

All the elements beyond uranium, the transuranium elements, have been prepared by bombardment and are not naturally occurring elements. The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle. Part A Complete the following equation: 10n+23892U→?+?01n+92238U→?+? Express your answer as a nuclear equation.

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Answer:

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Explanation:

Key statement;

The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle.

This is the beta particle; ⁰₋₁e

¹₀n+ ²³⁸₉₂U → Np + ⁰₋₁e

The mass number of Np;

1 + 238 = Np + 0

Np = 239

The atomic number of Np;

0 + 92 = Np + (-1)

92 + 1 = Np

Np = 93

The equation is given as;

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Which is not the name of a family on the periodic tablea) Halogens
b) Noble Gases
c) Alkali Earth Metals
d) Actinides

Answers

Answer:

I think it's D

Explanation:

Actinides is the correct answer
your noble gases are in the 8th column
Your Halogens are your 7th column
and you alkali earth metals are the 2nd column

What change would you expect on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion if the nucleophile concentration is halved and the alkyl halide concentration is unchanged ?