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3 years ago

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A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

esistance) the ball will land on the ground with a speed of
A. 0 m/s. B. 20 m/s. C. 30 m/s. D. 40 m/s. E. There is not enough information to say.

2 answers:

ICE Princess25 [194]3 years ago

6 0

B is the answer please mark brainliest

uysha [10]3 years ago

3 0

The answer is not B.

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A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

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

A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu

Free_Kalibri [48]

Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) = 2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

3 0

3 years ago

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What do you do if you approach a railroad crossing with a crossbuck sign that has no lights or gates

miv72 [106K]

Proceed with caution before crossing

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4 years ago

A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the

yuradex [85]

Answer:

$\frac{dy}{dt}=1.2\frac{mi}{min}$

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

$tan\theta=\frac{y}{x}\\y=xtan\theta(1)$

Differentiating (1) with respect to time, we get:

$\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\$

$\frac{dx}{dt}=0$ since x is a constant value. Replacing:

$\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}$

5 0

3 years ago