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VashaNatasha [74]
4 years ago
6

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

Physics
1 answer:
Wewaii [24]4 years ago
3 0

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

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Two technicians are discussing testing MAP sensors. Technician A says that the MAP sensor should be replaced if anything comes o
irinina [24]

The statements from both Technicians A and B are correct.

Answer: Option C

<u>Explanation:</u>

A typical MAP sensors comprises with a ceramic or silicon wafers, sealed with an ideal vacuum on one side and a suction manifold on the other. When the engine (motor) vacuum varies, the differential pressure across the board changes the output voltage or frequency to the MAP sensor. So, sensor vacuum to be increased if injection pulse widths increase.

Most pressure sensors operate at 5 volts from a computer and return a signal (voltage or frequency) based on the pressure applied to the sensor (vacuum). When testing the MAP sensor, make sure that the vacuum hose and hose connections are tightly connected to the engine vacuum source. According to this, concluding that the statements from both technicians are correct.

6 0
4 years ago
Which scenario is the best example of matter absorbing energy from a wave?
julia-pushkina [17]

Answer:

C I think so

Explanation:

ocean waves hitting a concert barrier that reduces their energy

7 0
3 years ago
Read 2 more answers
Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc
oee [108]

Answer:

The value is \Delta  \phi   =   4.12 \ rad

Explanation:

From the question we are told that

    The frequency of each sound is  f_1 = f_2 = f =  540 \  Hz

      The speed of the sounds is  v = 330 \  m/s

       The  distance of the first source from the point considered is  a = 4.40 \  m

        The distance of the second source from the point considered is  b  = 4.00  \  m

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_a =  2 \pi [\frac{a}{\lambda}  + ft]

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_b =  2 \pi [\frac{b}{\lambda}  + ft]          

Here b is the distance o f the first wave from the considered point  

Gnerally the phase diffencence is mathematically represented as  

           \Delta \phi= \phi_a - \phi_b  =  2 \pi [\frac{ a}{\lambda}  + ft ] - 2 \pi [\frac{b}{\lambda}  + ft ]      

=>      \Delta  \phi   =   \frac{2\pi [ a - b]}{ \lambda }

Gnerally the wavelength is mathematically represented as

        \lambda  =  \frac{v}{f}

=>     \lambda  =  \frac{330}{540}

=>     \lambda  =  0.611 \ m

=>    \Delta  \phi   =   \frac{2* 3.142 [ 4.40 - 4.0 ]}{  0.611  }

=>    \Delta  \phi   =   4.12 \ rad

     

5 0
3 years ago
If the period of a spring is 5 seconds what is the frequency
Scilla [17]

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

f=\frac{1}{T}

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

f=\frac{1}{5 s}=0.2 Hz

7 0
4 years ago
Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri
Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

E = \frac{Q}{4\pi \varepsilon_o r^2}

where;

  • E is the electric field
  • Q is the charge
  • r is the radius

The electric field reduces by a factor of \frac{1}{r^2}

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

A = 4\pi r^2

<h3>Change in area with r</h3>

\frac{dA}{dr} = 8\pi r

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0
2 years ago
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