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3 years ago

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a firework rock is burning at the rate of 14g/s all the products of combustion are ejected in one direction at the speed of 210m

/s what is the greatest height the rockect can have if it is moving vertically upward Give your answer in newtons

1 answer:

Naddika [18.5K]3 years ago

5 0

SO MY NAME IS ROHITH AND I LIKE MM

Explanation:

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A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

3 years ago

balu736 [363]

Answer:

B

Explanation:

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2 years ago

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Which symbol is used for an alpha particle? a. β c. γ b. Δ d. α

xenn [34]

It is the very last answer

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3 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

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3 years ago

Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en

allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

4 0

3 years ago