All categories

2 years ago

You went 100 miles north and 14

Physics

1 answer:

skad [1K]2 years ago

5 0

Answer:

82 degrees

Explanation:

consider your staying point to be the center of a circle. this center has the coordinates (0, 0).

the radius of the circle is the distance you walked East (14 miles).

I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.

then the 100 miles North is tan(displacement angle)×14.

as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.

so, again.

100 = 14×tan(angle)

tan(angle) = 100/14 = 50/7 = 7.142857...

the displacement angle is then 82 degrees.

You might be interested in

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago

Can any ideal gas power cycle have a thermal efficiency greater than 55 percent when using thermal energy reservoirs at 627∘C at

Yanka [14]

Answer:

Since the maximum thermal efficiency is higher than 55 percent, there can be a power cycle with these reservoir temperature with an efficiency higher than 55 percent.

Explanation:

The maximum thermal efficiency is determined from the given temperature

nth Carnot = 1- TL/TH

Where TL= 17+273= 290k

TH= 627*273= 900K.

nth Carnot = 1- 290/900 = 0.68

0.68*100 = 68 percent

8 0

2 years ago

What two states of matter are pictured in the image below?

Anuta_ua [19.1K]

Answer:

volume liquid

Explanation:

6 0

2 years ago

a string with fixed ends is made to go into standing wave patterns. at a given tension, the lowest frequency for which a certain

lakkis [162]

Answer:

700 hz

Explanation:

3 0

2 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. It keeps a space at 248C by consum-ing 2.15 kW of power. Det

Marizza181 [45]

Answer:

Heat of the reservoir is 461.38 K or 188.1 °C

The heating load is 18.705 kW

Explanation:

COP = 8.7

working temperature $T_{h}$ = 248 °C = 248 + 273.3 = 521.3 K

work power W = 2.15 kW

reservoir temperature $T_{c}$ = ?

heating load Q = ?

We know that

COP = Q/W

Q = COP x W = 8.7 x 2.15 = 18.705 kW

Also,

COP = $\frac{T_{h} }{T_{h}- T_{c} }$ = $\frac{521.3}{521.3- T_{c} }$

8.7 = $\frac{521.3}{521.3- T_{c} }$

4535.31 - 8.7 $T_{c}$ = 521.3

4535.31 - 521.3 = 8.7 $T_{c}$

4014.01 = 8.7 $T_{c}$

$T_{c}$ = 4014.01/8.7 = 461.38 K

or 461.38 -273.3 = 188.1 °C

3 0

3 years ago