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SCORPION-xisa [38]
3 years ago
15

Which organism in the food chain is sometimes first level consumer and sometimes 2nd level

Physics
1 answer:
makkiz [27]3 years ago
7 0
I believe the answer is Mouse/Herbivore. It was a question in my old lab haha.

Hope this helped!!
~xox Melis
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A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

Acceleration = (-28 m/s) / (13 sec)

Acceleration = -2.15 m/s²

5 0
3 years ago
Convert 435.5 lbs to kilograms
sergey [27]
Final Answer is - 197.53948
8 0
3 years ago
The ransom note in the Lindbergh kidnapping case included some obvious spelling errors and other foreign and regional language.
Dimas [21]

Answer: The FBI used a technique which is called Latent Fingerprint Identification: a silver nitrate solution was used to cause a reaction of the unseen fingerprints perspiration left by the kidnappers all over the ransom notes. The reaction caused the perspiration to turn into silver chloride. This substance is white and it is also visible to the naked eye which means the fingerprints of the perpetrators were easily identifiable.

sorry if this is wrong, i hope i helped

8 0
3 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi
yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

V=\frac{v^2m_e}{2e}

Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

8 0
3 years ago
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