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Mandarinka [93]

3 years ago

6

5. A family of ducks is swimming in a pond at a speed of 3 m/s when a gust of wind hits them. By the time they reach the other s

ide of the lack, the ducks are moving at a speed of 7 m/s. How long did it take for them to cross the lake if they accererated at a rate of 1 m/s2

1 answer:

ella [17]3 years ago

6 0

Answer:

The time taken by the duck to cross the lake is, t= 4 s

Explanation:

Given data,

The initial speed of the ducks, u = 3 m/s

The final speed of the ducks, v = 7 m/s

The acceleration of the duck, a = 1 m/s²

The formula for the acceleration is,

a = (v - u) / t

∴ t = (v - u) / a

Substituting the given values in the above equation,

t = (7 - 3) / 1

= 4 s

Hence, the time taken by the duck to cross the lake is, t= 4 s

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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

PLEASE HELPPPPP

Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (<em>zero movement</em>) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

<em>graph B</em>, and

<em>graph D</em>.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0

2 years ago

Yuliya22 [10]

PE= 3kg x 10N/kg x 10m
= 300J

8 0

3 years ago

A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal

SpyIntel [72]

Answer:

$x=1.75m$

Explanation:

From the exercise we have that

$y_{o}=1.3m\\v_{o}=3m/s, \beta =37\\$

<em><u>To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first </u></em>

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0m

$0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=-0.36 s$ or $t=0.73s$

Since the <u>time</u> can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

$x=v_{ox}t=3cos(37)(0.73)=1.75m$

6 0

3 years ago

Make the following conversion. 275 mm = _____ cm

scZoUnD [109]

10mm=1cm So.......... 275 mm=27.5 cm

6 0

3 years ago

Read 2 more answers