Activity that uses 150 calories of energy per day, or 1,000 calories per week, describes __________

san4es73 [151]

Answer: Hi!

A neuron is a basic working unit of the brain. Neurons are special cells designed to transfer information to other nerve, muscle, or gland cells. They are pretty cool - looking too! (A slightly irregular circular shape with branches reaching out from all sides.) A neuron is a nerve cell. Nerve cells are the way of communication in the nervous system.

Hope this helps!

3 0

2 years ago

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

so

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

5 0

2 years ago

I need to know what the answer is to this

Arturiano [62]

A pull between two objects, for example, between an object and Earth. When forces on an object are balanced, there is no change in speed or direction.

So the answer is I agree

7 0

1 year ago

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A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 66 t, at which time

musickatia [10]

(x=yt-2)^2(56-tea)^2x10r

5 0

2 years ago

A duck swimming on the surface of a pond has an

Gemiola [76]

From the given information:

Taking the movement of the Duck in the North as the x-direction
The movement of the Duck in the East direction as the y-direction

However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

$\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}$

The acceleration is:

$\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}$

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) = 4 seconds.

Using the first equation of motion:

$v^{\to} = u ^{\to} + a^{\to} t$

Then, we can replace their values into the equation of motion in order to determine the speed:

$\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}$

$\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}$

$\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}$

Collect like terms:

$\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0 )\hat x+( 0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}$

$\mathbf{v^{\to} =0.87500 \hat x- 1.01629 \hat y}$

Thus, the magnitude is:

$\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}$

$\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}$

$\mathbf{v^{\to} =\sqrt{1.79848}}$

$\mathbf{v^{\to} =1.34 \ m/s}$

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

brainly.com/question/17108011?referrer=searchResults

4 0

2 years ago

Activity that uses 150 calories of energy per day, or 1,000 calories per week, describes ___________.