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3 years ago

15

A space shuttle burns fuel at the rate of 13,000kg in each second. Find the force exerted by the fuel on the shuttle if in 2s th

e shuttle experiences a change in momentum of 325,000kgm/s.

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

162,500

Explanation:

P=F(t)

32500=F(2)

162,500

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What is the correct path of electrons?

MrRa [10]

Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.

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2 years ago

La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

Lina20 [59]

Answer:

Frequency, $f=4.41\times 10^{14}\ Hz$

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, $\lambda=6.8\times 10^{-7}\ m$

Speed of light is, $c=3\times 10^8\ m/s$

We need to find the frequency of visible red light. It can be calculated using below relation.

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz$

So, the frequency of visible red light is $4.41\times 10^{14}\ Hz$ .

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2 years ago

What is the correct answer?

LUCKY_DIMON [66]

Answer:

I think is d and you or very pretty

Explanation:

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2 years ago

What is another name for the introitus? psych 230?

Vera_Pavlovna [14]

Vaginal opening. areola is the part of the breast.

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2 years ago

A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.

Anna11 [10]

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

$B = \frac{\mu_{0}NI}{L}$

$\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m$

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

$N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns$

Since each turn has length 2πr of wire, the total length is:

$L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m$

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

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3 years ago