What happens to your image as you walk away from a plane mirror on a wall? It gets smaller.

A ball is thrown vertically upwards with a velocity

zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

6 0

2 years ago

Which of the following situations would cause the greatest decrease in the motion of molecules in a system?

Alina [70]

A. is the right answer since work is negative and Q which is heat in negative also

7 0

2 years ago

Read 2 more answers

A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the

Shkiper50 [21]

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

$K.E=\frac{1}{2}mv^2$

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)

v = velocity

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

$K.E=\frac{1}{2}mv^2$

$6.8J=\frac{1}{2}\times 0.046kg\times v^2$

$6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2$

$v=17.19m/s\approx 17m/s$

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

3 0

3 years ago

Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta

tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

$V = \frac{4}{3}\pi R_o^3* A$

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

So from above we can say that density of all nuclei is almost same.

5 0

3 years ago

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2

igor_vitrenko [27]

First, balance the reaction:

_ KClO₃ ==> _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃ ==> 2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃ ==> 2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0

2 years ago