In what way does the article contrast small cars with large cars?

What is the first step for both loading and unloading a firearm?

tekilochka [14]

Point the muzzle in a safe direction.

6 0

3 years ago

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

8 0

3 years ago

A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction)

klasskru [66]

Answer:

Explanation:

(a)

From the given information:

The initial velocity $v_1$ = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = $v_1 \ cos \ \theta$

$v_{1 \ x } = 5 \ cos \ 30^0$

$v_{1x} = 4.33 \ m/s$

The component along the y-axis = $v_2 { \ sin \ \theta}$

$v_{1 \ y } = 5 \ sin \ 30^0$

$v_{1 \ y } = 2.5 \ m/s$

To find the final velocity( reflected velocity)

using the same magnitude $v_2 = 5 \ m/s$

The angle from the x-axis can be $\theta_r = 90^0+60^0$

= 150°

Thus, the component along the x-axis = $v_2 \ cos \theta _r$

$v_{2x} = - 0.433 \ m/s$

The component along the y-axis = $v_2 \ sin \theta_r$

$v_{2y} = 5 \ sin \ 150^0$

$v_{2y} = 2.5 \ m/s$

(b)

The velocity $v_1$ can be written as in vector form.

$v_1 ^{\to} = v_1 x \hat {i} + v_1 y \hat {j}$

$v_1 ^{\to} =4.33 \ \hat {i} + 2.5 \ \hat {j}$ ---- (1)

The reflected velocity in vector form can be computed as:

$v_2 ^{\to} = v_2 x \hat {i} + v_2 y \hat {j}$

$v_2 ^{\to} =-4.33 \ \hat {i} + 2.5 \ \hat {j}$ --- (2)

The change in velocity = $v_2 ^{\to} - v_1 ^{\to}$

$\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j$

$\Delta v ^{\to} = - 8.66 \hat { i }$

(c)

The magnitude of change in velocity = $| \Delta V |$

$| \Delta V |$ = 8.66 m/s

6 0

3 years ago

What are two elements involved in the suns production of energy

skad [1K]

I know that radon is one

4 0

3 years ago

g Warm water in a geothermal heating system enters the pipe of a radiator at 20 psia and 119oF with a flow rate of 235 cfm (ft3/

Vika [28.1K]

Answer:

See explanation

Explanation:

Notice that the condenser section includes both the hot water and space heater and station (3) is specified as being in the Quality region. Assume that 50°C is a reasonable maximum hot water temperature for home usage, thus at a high pressure of 1.6 MPa, the maximum power available for hot water heating will occur when the refrigerant at station (3) reaches the saturated liquid state. (Quick Quiz: justify this statement). Assume also that the refrigerant at station (4) reaches a subcooled liquid temperature of 20°C while heating the air.

Using the conditions shown on the diagram and assuming that station (3) is at the saturated liquid state

a) On the P-h diagram provided below carefully plot the five processes of the heat pump together with the following constant temperature lines: 50°C (hot water), 13°C (ground loop), and -10°C (outside air temperature)

b) Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the P-h diagram.

c) Determine the mass flow rate of the refrigerant R134a. [0.0127 kg/s]

d) Determine the power absorbed by the hot water heater [2.0 kW] and that absorbed by the space heater [0.72 kW].

e) Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C [105 minutes].

f) Determine the Coefficient of Performance of the hot water heater [COPHW = 4.0] (defined as the heat absorbed by the hot water divided by the work done on the compressor)

g) Determine the Coefficient of Performance of the heat pump [COPHP = 5.4] (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)

h) What changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating

4 0

3 years ago