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4 years ago

Where does the cell store DNA and RNA

1 answer:

7 0

DNA is found mainly in the nucleus of the cell, while RNA is found mainly in the cytoplasm of the cell although it is usually synthesized in the nucleus.

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A motor-cycle rider going 90km/hr around a curve with a radius of 100m must lean at an angle to the vertical. The angle at which

BartSMP [9]

Answer:

The angle of inclination of the rider to the vertical is 32.5⁰.

Explanation:

Given;

linear speed of the motor cycle, v = 90 km/hr = 25 m/s

radius of the circle, r = 100 m

The angle of inclination of the rider to the vertical is calculated using banking angle formula;

$\theta = tan^{-1}(\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{25^2}{100 \times 9.8} )\\\\\theta = 32.5 ^0$

Therefore, the angle of inclination of the rider to the vertical is 32.5⁰

3 0

3 years ago

A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o

Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

Initial energy

$KE_i=\dfrac{1}{2}mu^2$

$KE_i=\dfrac{1}{2}0.3\times 10^2$

KEi= 15 J

Final kinetic energy

$KE_f=\dfrac{1}{2}mv^2$

$KE_f=\dfrac{1}{2}0.3\times 5^2$

KEf=3.75 J

The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

The minimum value to complete the circular arc

$V=\sqrt{r.g}$

Now by putting the values

$V=\sqrt{0.8\times 10}$

V= 2.82 m/s

So kinetic energy KE

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}0.3\times 2.82^2$

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

= 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0

3 years ago

A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?

AnnyKZ [126]

To answer this question, it helps enormously if you know
the formula for momentum:

Momentum = (mass) x (speed) .

Looking at the formula, you can see that momentum is directly
proportional to speed. So if speed doubles, so does momentum.

If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.

6 0

3 years ago

Read 2 more answers

Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of

iogann1982 [59]

Answer:

Explanation:

Momentum conservation

$m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\$

Kinetic energy conservation

$\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2$

Solve the system

6 0

4 years ago

Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to

stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

$\mu_{sB}=0.126$

$\mu_{sC}=0.168$

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

$\sum \tau_{A}=0$

$N(3m)-W(1.5m)=0$

When solving for N we get:

$N=\frac{W(1.5m)}{3m}$

$N=\frac{(1962N)(1.5m)}{3m}$

$N=981N$

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

$\sum F_{y}=0$