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3 years ago

Speed is a measure of how fast something moves in a specific direction true or false?

1 answer:

3 0

Speed is not a measure of how fast something moves <u>in</u> a specific direction. It is just the average speed per time unit without a direction. A direction is for velocity. The answer is false.
I hope this helps. :)

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Subtracting a negative number follows the same process as a positive number

Sav [38]

This would be true!

When you subtract a positive number, it goes to the left of the number line. When you subtract a negative number, it goes to the left of the number line.

The thing that confuses most people is that the first number (when negative) continues to increase as a negative. It follows the same process, just different signs.

I hope this helps!
~cupcake

6 0

3 years ago

H o m o zygous what does it mean

lord [1]

Answer:

two identical

Explanation:

6 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

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

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

Infuse 1000 mL of NS. The infusion set delivers 15 gtt/mL. Give 300 mL bolus over 20 minutes and then infuse at the hourly rate

Bas_tet [7]

Answer:

rate of infusion is 900 mL/hr

Explanation:

given data

Infuse I1 = 1000 mL

delivers = 15 gtt/mL

Infuse I2 = 300 mL

time t= 20 min

rate = 60 mL/hr

to find out

rate of infusion

solution

we know here we give 300 mL infuse in 20 min

so here for 20 min

rate of infusion is express as

rate of infusion = I2 / t

rate of infusion = 300 / 20

rate of infusion = 15 mL / min

rate of infusion = 15 × 60 = 900

so rate of infusion is 900 mL/hr

4 0

3 years ago

An air bubble has a volume of 1.70 cm³ when it is released by a submarine 115 m below the surface of a lake. What is the volume

Natalija [7]

Answer:

V = 20.67 cm³

Explanation:

In this case, let's apply the Boyle's law which is:

P1V1 = P2V2

Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.

By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²

We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:

P1 = Po + dgh

Where:

Po: innitial pressure, which we can assume is 1 atm

d = density of the substance, in this case, water (1000 kg/m³)

g = gravity (9.8 m/s²)

h = distance of the bubble from the surface (115 m)

Now replacing this data in the boyle's law we have the following:

P1V1 = P2V2

V2 = P1V1/P2

V2 = (Po + dgh) * V1 / P2

Replacing the data we have:

V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5

V2 = 2,087,600 / 1.01x10^5

V2 = 20.67 cm³

5 0

4 years ago