All categories

3 years ago

Plzzzz help me with this!!!!?????

Physics

2 answers:

Vikentia [17]3 years ago

4 0

A neap tide seven days after a spring tide

zepelin [54]3 years ago

3 0

Neap tide is when sun and moon are aligned at 90 degrees
spring tide is when sun and moon are in line 180 degrees.

You might be interested in

What is a precipitate?

Lunna [17]

Rain is considered a precipitate. Along with: drizzle, sleet, snow and hail.

hope this helps :)

7 0

3 years ago

Read 2 more answers

A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun

melomori [17]

Constant = straight line
“Travels at constant negative acc.”
Which is negative slope

Solution: B. Straight line w/ neg. slope

7 0

2 years ago

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

Fruits and vegetables should take up half you plate at a meal.<br> A. <br> True<br> B. <br> False

umka21 [38]

Answer:

A. true

Explanation:

3 0

2 years ago

Read 2 more answers

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0

3 years ago

Read 2 more answers