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Aleks [24]
3 years ago
7

PLEASE!!!!!!HELP!!!!!!!

Physics
1 answer:
REY [17]3 years ago
6 0
DNA Polymerase proofreads and corrects bases added during replication.
You might be interested in
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N
kotegsom [21]

Answer:

(A) = 34,560 J

(B) Work done during the round trip is not zero, hence the resistive force is not a conservative force.

Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

SOLUTION:

distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fe) = 3.3 N due east

(A) to find the work done by the resistive force during the round trip we have to get the work done by the resistive force due west and add it to the work done by the resistive force due east

  • work done by the resistive force due west = (Fw.cosθ) x De

        where θ is the angle between the displacement and the force. The    

        displacement is due east while the force is due west, hence θ = 180°

       work done (W1) = (3.1 x cos 180) x 5400 = -16,740 j

  • work done by the resistive force due east = (Fe.cosθ) x Dw

        where θ is the angle between the displacement and the force. The    

        displacement is due west while the force is due east, hence θ = 180°

       work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

Hence work done during the round trip = W1 + W2 = (-16,740 ) + (-17,820)

= 34,560 J

(B) A conservative force is a force in which the work done by the force in moving an object around a closed path is zero. From part (A) above the work done during the round trip is not zero, hence the resistive force is not a conservative force.

3 0
3 years ago
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
4 years ago
A 10 kg block slides on a frictionless surface at 10 m/s . It hits a rough patch 3 meters long with a coefficient of kinetic fri
Valentin [98]

12 MPH

I DIDNT do the math my brother did  hes in colledge so good lucks guys

8 0
3 years ago
A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?
Luden [163]

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

3 0
3 years ago
Read 2 more answers
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