All categories

4 years ago

Which statement best describe nucleic acids

Physics

2 answers:

Phantasy [73]4 years ago

6 0

Answer:

Where are the statements

Explanation:

Elis [28]4 years ago

6 0

Nucleic acids are the biopolymers which are essential to all the life

You might be interested in

Friction and normal force are examples of

siniylev [52]

Answer:

Friction is a force that tends to oppose the relative motion between two bodies in contact. Frictional force always acts on a moving body from the direction opposite to the direction of motion. It opposes the motion, and therefore, helps to reduce the speed of the moving object. It is a contact force.

3 0

2 years ago

A coil consists of 180 turns of wire. Each turn is a square of side d=30 cm, and a uniform magnetic field directed perpendicular

alexira [117]

Answer:

Emf induced i equal to 329.4 volt

Explanation:

Note : Here i think we have to find emf induced in the coil

Number of turns in the coil N= 180

Sides of square d = 30 cm = 0.3 m

So area of the square $A=0.3\times 0.3=0.09m^2$

Magnetic field is changes from 0 to 1.22 T

Therefore $dB=1.22-0=1.22T$

Time interval in changing the magnetic field dt = 0.06 sec

Induced emf is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-180\times 0.09\times \frac{1.22}{0.06}=329.4volt$

8 0

4 years ago

Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 2.9 kg pumpkin to a height of 1.5 m

Katen [24]

Answer:

Work done in lifting to a height of 1.5 m is 42.63 J.

Work done in carrying it to 50 m is 0 J

Explanation:

Given:

Mass of the pumpkin (m) = 2.9 kg

Vertical displacement of the pumpkin (y) = 1.5 m

Horizontal displacement of the pumpkin (x) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done by a force is given by the formula:

$W=FS\cos\theta$

Where,

$F\to force\ applied\\\\S\to displacement\ caused\\\\\theta\to angle\ between\ F\ and\ S$

Now, the work done is maximum when both force and displacement is in same direction. ( $\theta=0^\circ$ )
Work done is minimum when both force and displacement are in opposite direction. ( $\theta=180^\circ$ )
Work done is zero when force and displacement are perpendicular to each other. ( $\theta=90^\circ$ )

Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.

So, work done is given as:

$W=Force\times Vertical\ displacement$

Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.

So, Force = mg = $2.9\times 9.8 = 28.42\ N$

So, work, $W=Fy=28.42\times 1.5=42.63\ J$

So, work done in lifting the pumpkin is 42.63 J.

Now, when the pumpkin is carried to the check-out stand, the displacement is horizontal while the force applied is still in the upward direction.

So, the force and displacement are perpendicular to each other. Thus, the work done is zero.

4 0

4 years ago

HELP PLZZZZZ NOW!!!

Aneli [31]

Answer:Climatic processes affect the dynamics of Earth's ice sheets and glaciers, and along ... by abrupt events and by continuous reshaping of Earth's surface from surface ... Forecasting natural disasters, including the timing and size of earthquakes, the . Last, human activity has a profound impact on water resources, landscape

Explanation:

step by step

3 0

3 years ago

Read 2 more answers

A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The

gregori [183]

First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.

Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:

$\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}$

This equation reads:

The velocity of the arrow with respect to the ground is equal to the velocity of the arrow with respect to the bow plus the velocity of the bow with respect to the gound.

Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:

$\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}$

Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:

$\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}$

$\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}$

Similarly, for the velocity of the bow with respect to the ground:

$\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}$

Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:

$\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}$

Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.

Since the horizontal movement of the arrow is uniform, then:

$v_{AG-x}=\frac{x}{t}_{}$

Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:

$\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}$

The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:

$y=v_{AG-y}t-\frac{1}{2}gt^2$

Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:

$\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}$

Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.

6 0

1 year ago