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3 years ago

The missile s engine exerts a 260-kN force F.

Physics

1 answer:

alexdok [17]3 years ago

6 0

Answer:

a)F=(208 i+156 j) kN

b)F+W=219 kN

Explanation:

Solution:

To determine the components of F

F=(260 kN).( $\frac{4}{\sqrt{4^{2}+3^{2}}}i+\frac{3}{\sqrt{4^{2}+3^{2}}}j$ )

=(208 i+156 j) kN

F=(208 i+156 j) kN

b)weight W expressed in component and add into the

Force F.

W=-(8800 kg).(9.81 m/s^2)j

F+W=(208 i+[156-86.3]j) kN

F+W= $\sqrt{(208kN)^{2}+(-69.7kN)^{2} }$

=219 kN

F+W=219 kN

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A wooden beam with a mass of 10.8 kg is supported by vertical supports

blondinia [14]

Answer:

Force from the support closest = 79.8 N

Force from the support furthest = 61.9 N

Explanation:

Let's say the length of the beam is L. Let's say A is the near end of the beam and B is the far end of the beam.

Draw a free body diagram. There are four forces on the beam:

Reaction force Ra at the near end (0),

Reaction force Rb at the far end (L),

Weight force of the beam Mg at the center (L/2),

Weight force of the book mg at L/4 from A.

Sum of torques at A:

∑τ = Iα

Rb (L) − Mg (L/2) − mg (L/4) = 0

Rb (L) = Mg (L/2) + mg (L/4)

Rb = ½ Mg + ¼ mg

Rb = (½ M + ¼ m) g

Rb = (½ (10.8 kg) + ¼ (3.66 kg)) (9.8 m/s²)

Rb = 61.9 N

Sum of forces in the y direction:

∑F = ma

Ra + Rb − Mg − mg = 0

Ra = Mg + mg − Rb

Ra = (M + m) g − Rb

Ra = (10.8 kg + 3.66 kg) (9.8 m/s²) − 61.9 N

Ra = 79.8 N

7 0

3 years ago

Which geologic events occur most often at this mid-oceanic ridge plate

Leni [432]

Answer: Earthquakes and Volcanic eruptions

Explanation:

You did not attach a picture however, this should be the correct answer.

Mid-oceanic ridge plate boundaries such as the Mid-Atlantic Ridge form as a result of plate separating at the boundary. This leads to earthquakes due to the forces of tension acting on the plates.

When this happens, the mantle is exposed and therefore there will be volcanic eruptions as the lava comes to the surface as magma. When this magma cools it creates new seafloors and underwater mountains.

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3 years ago

Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show

ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = $\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}$

We know that the inductance of the coil is given by the formula:

$L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH$

4 0

3 years ago

What are the two categories of waves

fredd [130]

answer: transverse and longitudinal

6 0

3 years ago

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago