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3 years ago

Can someone please help me with questions 5 and 6 you would be so amazing! :)

Physics

1 answer:

gogolik [260]3 years ago

4 0

Answer: 5. Adenosine triphosphate (ATP) consists of an adenosine molecule bonded to three phophate groups in a row. In a process called cellular respiration, chemical energy in food is converted into chemical energy that the cell can use, and stores it in molecules of ATP. This occurs when a molecule of adenosine diphosphate (ADP) uses the energy released during cellular respiration to bond with a third phosphate group, becoming a molecule of ATP. So the energy from cellular respiration is stored in the bond between the 2nd and 3rd phosphate groups of ATP. When the cell needs energy to do work, ATP loses its 3rd phosphate group, releasing energy stored in the bond that the cell can use to do work. Now its back to being ADP and is ready to store the energy from respiration by bonding with a 3rd phosphate group. ADP and ATP constantly convert back and forth in this manner.

6. Active transport directly uses a source of chemical energy (e.g., ATP) to move molecules across a membrane against their gradient.

explanation: simplify 5. in your own words.

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A charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units. Where, along the x-axis is the net force on a charge

RideAnS [48]

At x = 1.72 units the net force on a charge Q equal to zero when a charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units.

A net force is defined as the sum of all the forces acting on an object. The equation below is the sum of N forces acting on an object. There may be several forces acting on an object, and when you add up all of those forces, the result is what we call the net force acting on the object.

This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

+3q ⇔ +11q

x = ? ⇔ x = 5

$\frac{1}{4\pi E0} (\frac{3q}{x^{2} } -\frac{11q}{(5-x)^{2} } ) = 0$

$\frac{3}{x^{2} } = \frac{11q}{(5-x)^{2} }$

$\frac{5-x}{x} = \sqrt{\frac{11}{3} } = 1.91\\$

$5-x = 1.91x\\x = 1.715 = 1.72 units$

∴ At x = 1.72 units the net force on a charge Q equal to zero

Learn more about net force here

brainly.com/question/14063083

#SPJ4

5 0

11 months ago

At the end of a bike ride up a mountain, Chris was at an elevation of 500 meters above where he started. If Chris's mass is 60kg

lawyer [7]

294000J

Explanation:

Given parameters:

Height of hill = 500m

Mass of Chris = 60kg

Unknown:

Potential energy increase = ?

Solution:

Potential energy is the energy by virtue of the position of a body.

Potential energy = mgh

m is the mass of the bike

g is the acceleration due to gravity

h is the height

At the base of the hill,

P.E = mgh = 60 x 9.8 x 0 = 0J

At the top:

P.E = mgh = 60 x 9.8 x 500 = 294000J

Potential energy increase = 294000J - 0J = 294000J

learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

8 0

3 years ago

How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c

romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

$V =\frac{W}{\gamma }$

Here, W is the weight of concrete canoe and $\gamma$ is the specific weight of water and its value is $62.4\ lb/ft^3$ .

So,

$V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3$ .

Now the volume of water occupied in ultra lightweight kevlar canoe,

$v=\frac{w}{\gamma}$

Here, w is weight of kevlar canoe.

So,

$v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3$

Thus, the volume of water displaced,

$=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3$ .

Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is $2.19\ ft^3$

5 0

2 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

13 points and brainlyest if possible. Thanks.

nikdorinn [45]

Most likely it would be C not completely sure

3 0

2 years ago

Read 2 more answers