In which part of the scientific method do you tell why the research is being done?

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass $m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}$

$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

3 years ago

How heavy is an object that displaces 400N of water in a pool?

julia-pushkina [17]

All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.

7 0

3 years ago