Ask question

Ask question

All categories

4 years ago

8

A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds

at an average acceleration of aav,x = -0.105 m/s2 and aav,y = 0.101 m/s2. What are the components of the speedboat\'s final velocity, v2,x and v2,y?
Find the speedboat's final speed?

1 answer:

KengaRu [80]4 years ago

3 0

First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s

You might be interested in

A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0

3 years ago

study a 75kg person stands on a scale in an elevator what does read in n and in kg when the elevator is at rest?

Verdich [7]

To solve this problem we will use Newton's second law in order to obtain the weight of a person. The second law tells us that

F = ma

Where,

m = mass

a = Acceleration

In this particular case, the acceleration is equal to that exerted by the earth through gravitational acceleration, so if the person's weight is 75Kg and the gravity is $9.8m / s^2$ , the weight of the body will be,

$F_W =mg$

$F_W = (75kg)(9.8m/s^2 )$

$F_W = 735.75N$

When the elevator is at rest this reads 735.75N and 75Kg (The same mass of the person)

6 0

3 years ago

if a water wave vibrates up and down 4 times each second,the distance between 2 successive crest is 5 meters, and the height fro

Mnenie [13.5K]

frequency=4Hz

wavelength=5m

amplitude=1/2×2=1m

period=1/frequency

1/4=0.25seconds.

velocity=wavelength×frequency

=5×4

=20m/s

5 0

3 years ago

Read 2 more answers

What causes a objects to move or stay still? claim and evidence

jek_recluse [69]

Answer:

A force

Explanation:

A push or a pull is an example of a force and can cause an object to speed up, slow down, etc.. Newton's laws tell us that 1- an object will not change its motion unless a force acts on it 2- the force on an object is equal to its mass times its acceleration. 3- The third law states that for every action (force) in nature there is an equal and opposite reaction.. However, forces like gravity and friction can resist movement.

4 0

3 years ago

Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have

Lera25 [3.4K]

Answer:

The extent of greenhouse effect on mars is $G_m = 87 K$

Explanation:

From the question we are told that

The albedo value of Mars is $A_1 = 0.15$

The albedo value of Mars is $A_2 = 0.15$

The surface temperature of Mars is $T_1 = 220 K$

The surface temperature of Venus is $T_2 = 700 K$

The distance of Mars from the sun is $d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m$

The distance of Venus from sun is $d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m$

The radius of the sun is $R = 7*10^{8} \ m$

The energy flux is $E = 6.28 * 10^{7} W/m^2$

The solar constant for Mars is mathematically represented as

$T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ]$

Where $\sigma$ is the Stefan's constant with a value $\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}$

So substituting values

$T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}$

$T = 307K$

So the greenhouse effect on Mars is

$G_m = T - T_1$

$G_m = 307 - 220$

$G_m = 87 K$

3 0

4 years ago