How much force is required to accelerate a 5 kg mass at 20 m/s^2

A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative

Liula [17]

Question is missing. Found on google:

a) What are the components of the acceleration of the fish?

(b) What is the direction of its acceleration with respect to unit vector î?

(c) If the fish maintains constant acceleration, where is it at t = 30.0 s?

(a) $(0.73, -0.47) m/s^2$

The initial velocity of the fish is

$u=(4.00 i + 1.00 j) m/s$

while the final velocity is

$v=(15.0 i - 6.00 j) m/s$

Initial and final velocity are related by the following suvat equation:

$v=u+at$

where

a is the acceleration

t is the time

The time in this case is t = 15.0 s, so we can use the previous equation to find the acceleration, separating the components:

$v_x = u_x + a_x t\\a_x = \frac{v_x-u_x}{t}=\frac{15.0-4.00}{15.0}=0.73 m/s^2$

$v_y = u_y + a_y t\\a_y = \frac{v_y-u_y}{t}=\frac{-6.00-1.00}{15.0}=-0.47 m/s^2$

(b) $-32.8^{\circ}$

The direction of the acceleration vector with respect to i can be found by using the formula

$\theta = tan^{-1}(\frac{a_y}{a_x})$

where

$a_x$ is the horizontal component of the acceleration

$a_y$ is the vertical component of the acceleration

From part a), we have

$a_x = 0.73 m/s^2$

$a_y = -0.47 m/s^2$

Substituting,

$\theta = tan^{-1}(\frac{-0.47}{0.73})=-32.8^{\circ}$

(c) $r=(460.5 i - 185.1 j )m$

The initial position of the fish is

$r_0 = (12.0 i -3.60 j) m$

The generic position r at time t is given by

$r= r_0 + ut + \frac{1}{2}at^2$

where

$u=(4.00 i + 1.00 j) m/s$ is the initial velocity

$a=(0.73 i -0.47 j) m/s^2$ is the acceleration

Substituting t = 30.0 s, we find the final position of the fish. Separating each component:

$r_x =12.0 + (4.00)(30) + \frac{1}{2}(0.73)(30)^2=460.5 m\\r_y = -3.60 + (1.00)(30) + \frac{1}{2}(-0.47)(30)^2=-185.1 m$

So the final position is

$r=(460.5 i - 185.1 j )m$

4 0

3 years ago

A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a

kirill [66]

Answer:

θ = 13.16 °

Explanation:

Lets take mass of child = m

Initial velocity ,u= 1.1 m/s

Final velocity ,v=3.7 m/s

d= 22.5 m

The force due to gravity along the incline plane = m g sinθ

The friction force = (m g)/5

Now from work power energy

We know that

work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2 g sinθ - ( 2 g)/5 ) d = v² - u²

take g = 10 m/s²

(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²

20 sinθ - 4 =12.48/22.5

θ = 13.16 °

5 0

3 years ago

Pepe and alfredo are resting on an offshore raft after a swim. they estimate that 3.00 m separates a trough and an adjacent cres

Orlov [11]

The velocity (V) of a wave is the frequency (F) times the wave length (lambda):

V = F * lamda

lambda is the distance from crest to crest which is twice the distance from crest to trough.

=> lamba = 2 * 3.00 m = 6.00 m

F = number of waves / time = 13.0 waves / 20.2 s

Now you can plug in the values in the formula of V:

V = 6.00 m/wave* 13.0 waves / 20.2 s = 3.86 m/s

Answer: 3.86 m/s

8 0

3 years ago

X + 10 times X = 50 what is the answer

tia_tia [17]

Answer:

4.54

Explanation:

X+10X=50

11X=50

X=4.54#

<h2>please follow me...</h2>

3 0

3 years ago

a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her ener

azamat

Answer:

belpw

Explanation:

7 0

3 years ago