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3 years ago

A person drives 4 miles to the mall. If the speed limit is 45 miles/hr and the person

Physics

1 answer:

tiny-mole [99]3 years ago

6 0

Answer: Third Option

Yes, the person is going 133 miles/hr.

Explanation:

Remember that the average speed can be calculated as

$s = \frac{d}{t}$

Where s is the speed

d is the distance traveled

t is the time

In this case

$d =4\ miles\\$

$t=0.03\ hours$

Therefore

$s = \frac{4}{0.03}$

$s = 133.33\ miles/hr$

Remember that the speed limit is 45 miles/hr

So the answer is the third option "Yes, the person is going 133 miles/hr".

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A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

$\Delta E_{system} = E_{in} - E_{out}$

$\Delta U = W_{in} - Q_{out}$

Hence, $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

$\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

$T_{2} = \frac{20 psia}{14.7}(638 R)$

= 868.03 R

Now, we will calculate the mass as follows.

m = $\frac{P_{1}V}{RT_{1}}$

= $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

= 1.031 lbm

Hence,

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

$W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

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6 0

3 years ago

Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th

Papessa [141]

<span>3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m</span>

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3 years ago

During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu

matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.