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german
2 years ago
5

Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so tha

t rocket slows down at the rate of 2.05 m/s2 . A 6.50-kg instrument is hanging by a vertical wire inside a space ship.Find the force that the wire exerts on the instrument.
Physics
1 answer:
Valentin [98]2 years ago
5 0

Answer:

R= 78.32 N

Explanation:

Given that

Acceleration ,a= 2.05 m/s²

Mass , m = 6.5 kg

The force due to acceleration

F= mass x Acceleration

F=  ma

F= 6.5 x 2.05 N

F= 13.32 N

The force due to weight

F' = m g

F' = 6.5 x 10 N             ( take g= 10 m/s²)

F'= 65 N

Therefore the net total force will be summation of force due to weight and force due to acceleration

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

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A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
irina [24]

Answer:

The atomic weight in g/mole of the metal (molar mass) is 8.87.

Explanation:

To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The chemical formula of metal oxide tells you that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

In the previous equation you can see that you need 3  oxygen anions to react with two metal cations. Then:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

You have  52.92 g  of metal in the sample, then the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>

The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
2 years ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 years ago
Capillary waves travel what than long waves
7nadin3 [17]
Faster than. Hope this helps!!!
6 0
2 years ago
Read 2 more answers
The speed of sound in air changes with the temperature. When the temperature T is 32 degrees Fahrenheit, the speed S of sound is
dezoksy [38]

Answer and Explanation:

A. We have temperature t = 32

Speed of sound, s = 1087.5

As t increases by 1⁰f speed increases by 1.2

So that

S = 1088.6

T= 33⁰f

We have 2 equations

1087.5 = k(32) + c

1088.6 = k(33) + c

Subtracting both equations

(33-32)k = 1088.6-1087.5

K = 1.1

b.). S = kT + c

1087.5 = 32(1.1) + c

Such that

C = 1052.3

Therefore

S = 1.1(t) + 1052.3

C.). S = 1.1t + 1052.3

We make t subject of the formula

T = s/1.1 - 1052.3/1.1

T = 0.90(s) - 956.3

D. This means that We have temperature to rise by 0.90 whenever speed is increased

8 0
2 years ago
a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll
marissa [1.9K]

Answer:

h = 0.0362\,m

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

U_{g,A} + K_{A} = U_{g,B} + K_{B}

U_{g,B} - U_{g,A} = K_{A} - K_{B}

m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}

The moment of inertia of the coin is:

I = \frac{1}{2}\cdot m \cdot r^{2}

After some algebraic handling, an expression for the maximum vertical height is derived:

m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}

h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}

h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

h = 0.0362\,m

3 0
2 years ago
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