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2 years ago

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Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh

ot horizontally with a speed of 8 m/s that lands 10 m away.

1 answer:

Salsk061 [2.6K]2 years ago

8 0

Answer:

a

Explanation:

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

$p = \frac{fd}{t} \\ where \: \\p = power \\ f = force \\ d = distance \\ and \: t = time \\ \\ p = \frac{3800 \times 50}{10} \\ p = \frac{190000}{10} \\ p = 19000w$

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3 years ago

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

denis-greek [22]

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$

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3 years ago

Which type of material uses the subtractive coloring process?

Triss [41]

I think that the answer is A

6 0

2 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

6 0

3 years ago