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2 years ago

5

Water is considered to be

1 answer:

Natalka [10]2 years ago

3 0

Answer: the thing that brings you life- h2o- idrk the question but if this helps your welcome ^-^have a good day

Explanation:

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A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

Ivahew [28]

<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>

5 0

3 years ago

As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat

DaniilM [7]

Answer:

the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

Explanation:

Given that :

$\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}= \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399$

$V = e^{kt} + 399$

When v = 300 ; $\frac{dv}{dt}= - 15$

then

$\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05$

∴ $\\V = e^{0.05t} + 399\\$

Therefore, the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

6 0

3 years ago

A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.

Kay [80]

Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g= $0.27\times 10^{-3} kg$

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm= $8.5\times 10^{-2} m$

1m=100 cm

a.Potential energy stored in the compressed spring is given by

P.E= $\frac{1}{2}kx^2$

$P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2$

$P.E=6.5025 J$

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0

3 years ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

3 0

1 year ago

The elememt sodium (Na) is a ?

luda_lava [24]

It is a very reactive metal with 11 protons ,12 neutrons, 11 electrons, and 1 valence electron

3 0

3 years ago

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