WHAT HAPPENS TO THE VELOCITY AND ACCELERATION OF A BODY FALLING FREELY IN A VACUUM?

Some steamboat restaurants use paper pots for their customers to boil the food themselves. What is the reason for the paper not

Ber [7]

Answer: D. 1,2,3

The paper pots used in steamboat restaurants will not catch fire when in contact with the flame for the heat is conducted quickly to the water because the paper is thin and the burning temperature of the paper is much higher than the boiling point of water. Lastly, the texture of the paper is thick that it can withstand the temperature of the flame.

7 0

3 years ago

Determine the number of atoms in the following chemical formulas. Make sure you list the elements, how many atoms of each elemen

Oksanka [162]

Answer:

Please mark as Brainliest!!

Explanation:

Explanation:

In a chemical formula, the symbols for each element in the compound are followed by subscripts that tell us how many of that element are in the compound.

For an example, let's look at the the formula for compound glucose:

C 6 H 12 O 6

________

The subscripts that follow each element's symbol indicate how many of that element are in the compound.

7 0

3 years ago

Read 2 more answers

The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^

allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

F = G m1 m2 / r^2

G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2

= 3820885 .3 N

6 0

2 years ago

K-2/5=11k literal equations

krek1111 [17]

-2/5 = 11k - k
-2/5 = 10k
-2/5/10 = k
-2/5 * 10 = k
-2/50 = k
k = -1/25.

-1/25 - 2/5 = 11k is true.

6 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$