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sergey [27]
2 years ago
10

A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1

.23 s? (Ignore direction)
Physics
1 answer:
aev [14]2 years ago
3 0

The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

u is the initial velocity of canonball= 47.4 m/sec

g is the acceleration of free fall = 9.81 m/sec²

v is the velocity after 1.23 s

According to Newton's first equation of motion,

\rm v=u +gt \\\\ v= 47.4 \  m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec

Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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If the wavelength of an s-wave is 23,000 m, and its speed 4500 m/s, what is its frequency?
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Answer:

5 metre.

Explanation:

Wavelength = Velocity / Frequency

= 23,000/ 4,500

= 5 metre.

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How strong would the gravitational attraction be (in units of Earth-pulls) between a Saturn-like planet and the Sun, if the plan
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8 0
3 years ago
A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at
kherson [118]

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity <em>u </em>and it falls under the acceleration due to gravity<em> g.</em>

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

s=ut+\frac{1}{2}at^2

The bag makes a net displacement <em>s</em> of 13.4 m downwards, hence

s=-13.4 m

Its initial velocity is

u=+4.6 m/s

The acceleration due to gravity acts downwards and hence it is negative.

g=-9.8 m/s^2

Use the values in the equation of motion and write an equation for t.

s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0

Solving the equation for t and taking only the positive value for t,

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4 0
3 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the h
Kazeer [188]

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

f=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

f_1 = rate constant at 525K

K_2 = rate constant at 545K

Ea = activation energy for the reaction = 185kJ/mol= 185000J/mol   (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 525 K

T_2 = final temperature = 545 K

Now put all the given values in this formula, we get

\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]

\log (\frac{f_2}{f_1})=0.6754

(\frac{f_2}{f_1})=4.736

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736

8 0
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3 0
3 years ago
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