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3 years ago

What was the most significant accomplishment of the apollo program?

2 answers:

7 0

The most significant accomplishment of the Apollo space program
was the clear public demonstration that the US has the rocket-booster
capacity to launch heavy nuclear warheads to any point on Earth, and
the navigational ability to place them precisely where we want them.

An incidental accomplishment of the program was a great boost to
US national pride by landing men on the moon for the first time ever,
and returning them safely to Earth.

Nitella [24]3 years ago

3 0

C.landing an astronaut on the moon.

You might be interested in

Surgical repair of a muscle

Misha Larkins [42]

A myoplasty is the surgical repair of a muscle.

6 0

4 years ago

A 15 kg block is on a ramp which is inclined at 20o above the horizontal. It is connected by a string to a 19 kg mass which hang

12345 [234]

Answer:

The magnitude of the acceleration of the 19 kg block is 1.414 m/s²

Explanation:

From Newton's second law of motion;

$F_{Net} = ma$

where;

m is the mass of the objects involved, kg

a is the acceleration of the object, m/s²

different forces on the block and string

⇒force due to 15 kg block

=mgcosθ = 15×9.8×cos20 = 15×9.8×0.9396

= 138.12 N

⇒Tensional Force on 19 kg mass:

T = mg = 19×9.8 = 186.2 N

$F_{Net} = T-mg = a(m_1+m_2)$

186.2 - 138.12 = a(15+19)

48.08 = a(34)

a = 48.08/34

a = 1.414 m/s²

Therefore, the magnitude of the acceleration of the 19 kg block is 1.414 m/s²

7 0

4 years ago

What is the correct answer?

dexar [7]

Answer:

length=2

Explanation:

You simply find the zeros of the quartic polynomial

The factored form is (x+1)(x-5)(x-2)=0

The zeros become -1,5 and 2

Dimension 5 is already given and the question says length>width and 2>-1

Therefore length=2

5 0

3 years ago

Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or

Stolb23 [73]

Answer:

I = (1.80 × 10⁻¹⁰) A

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = (1.0 × 10⁻¹⁵) T

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 3.6 cm = 0.036 m

(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)

4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036

I = (1.80 × 10⁻¹⁰) A

Hope this Helps!!!

3 0

4 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago