PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST

Which is an example of the force of attraction between two objects that have mass?

Genrish500 [490]

According to multiple scientific experiments, most objects develop an attraction towards Candice.

3 0

3 years ago

Question 2 Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid

Mama L [17]

Answer:

The answer is 12.27 g (NH4)3PO4

Explanation:

Step 1: balance the chemical equation:

H3PO4 + 3NH3 → (NH4)3PO4

step 2: the molar masses of each of the reagents and products are calculated using the periodic table:

H3PO4:

3 atoms of H: 3x1=3 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=3+30.97+64=97.97 g/mol

3NH3:

3 atoms of N: 3x14=42 g/mol

9 atoms of H: 9x1=9 g/mol

Molar mass=42+9=51 g/mol

(NH4)3PO4:

3 atoms of N: 3x14=42 g/mol

12 atoms of H: 12x1=12 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=42+12+30.97+64=148.97 g/mol

we make a rule of three to calculate the amount of ammonium phosphate:

51 g NH3------------------148.97 g (NH4)3PO4

4.2 g NH3-----------------x g (NH4)3PO4

Clearing the x, we have:

$x g (NH4)3PO4 = \frac{(4.2 g NH3)x(148.97 g (NH4)3PO4)}{51 g NH3}=12.27 g (NH4)3PO4$

7 0

3 years ago

When a distant star explodes, a visible light telescope and s radio telescope located on earth receive signals from the explodin

lorasvet [3.4K]

Answer:

the correct answer is D

Explanation:

When the star explodes the radiation travels through empty space at the speed of light c= 3 10⁸ m/s

This speed has been experimentally proven to be constant, therefore two two instruments arrive at the same time

Therefore the correct answer is D

4 0

3 years ago

Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid

algol13

Answer:

The intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

$|E_{1}|=k\frac{q_{1}}{d^{2}}$

Where:

k is Coulomb's constant
q(1) is 20.00 μC or 20*10⁻⁶ C
d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

$|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}$

$|E_{1}|=1.8*10^{7}\: N/C$

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

$|E_{2}|=k\frac{q_{2}}{d^{2}}$

$|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}$

$|E_{2}|=7.2*10^{6}\: N/C$

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

I hope it helps you!

5 0

3 years ago

A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m.

Aloiza [94]

When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.

Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration

$\frac{ M_{E}*G }{ ( r_{E} )^{2} } =$ ω^2r

where

$M_{E}$ = mass of earth
G = gravitational constant = 6.6742 x 10-11 m3 s-2 kg-1
 $r_{E}$ = radius of earth
ω = angular velocity
r = radius of orbit

To convert to angular velocity:

Tangential velocity = rω
ω = 5000/r

Then,

$\frac{ (6 \ x \ 10^{24}) *(6.6742 \ x \ 10^{-11} ) }{ ( 6.4 \ x \ 10^{6})^{2} }= ( \frac{5000}{r} )^{2} r$

r = 2557110.465 m

Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.

4 0

3 years ago

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