Which of the following is the correct formula for calculating power

What is the speed of an object if it travels 15m in 5seconds

Thepotemich [5.8K]

6.71081mph(miles per hour) or 3m/s (meters per second)

3 0

3 years ago

A small ball is hung by a string from the ceiling of a train car, so that the ball hangs vertically. Moving from rest, the train

kondor19780726 [428]

Answer:

m

Explanation:

5 0

4 years ago

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

PLS HELP

Ahat [919]

"This was the idea that non-living objects can give rise to living organisms."

7 0

3 years ago

What is a normal resting heart rate for adults, according to the American

kicyunya [14]

The answer of this question is
No B

7 0

3 years ago