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Elena-2011 [213]

2 years ago

12

a speed swimmer love to race around the parks pine which is 25m around if she can swim 20 laps in 7200s what is her average spee

d?

1 answer:

aleksley [76]2 years ago

8 0

Answer:

the average speed of the swimmer is 0.069 m/s.

Explanation:

Given;

complete distance around the park pine, d = 25 m

total lap completed, = 20 laps

time of laps completion, t = 7200 s

The total distance completed by the swimmer = 20 x 25 = 500 m

The average speed of the swimmer = distance / time

= (500 m) / (7200 s)

= 0.069 m/s.

Therefore, the average speed of the swimmer is 0.069 m/s.

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A parachute falling to the ground.<br><br>

butalik [34]

Answer:

a parachute falling to the ground is uniform

7 0

1 year ago

The initial speed of a body is 7.1 m/s. What is its speed after 2.23 s if it accelerates uniformly at 2.64 m/s 2 ? Answer in uni

Nana76 [90]

13.0m/s

1.2m/s

Explanation:

Given parameters:

Initial speed of the body = 7.1m/s

time taken = 2.23s

Acceleration = 2.64m/s²

Unknown:

Final speed = ?

Solution:

Acceleration is the rate of change of velocity with time.

a = $\frac{V - U}{T}$

a = acceleration

V = final speed

U = initial speed

T = time taken

Input the variables and solve for V;

2.64 = $\frac{V - 7.1}{2.23}$

V - 7.1 = 5.9 expression 1

V = 5.9 + 7.1 = 13.0m/s

B

Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;

from expression 1;

V - 7.1 = -5.9

V = -5.9 + 7.1 = 1.2m/s

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

2 years ago

A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho

Lemur [1.5K]

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

6 0

2 years ago

As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?

irga5000 [103]

The Charge Gets Smaller.

5 0

2 years ago