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ollegr [7]
3 years ago
7

Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2

has a mass 6.59 kg and is moving to the left at 10.4 m/s. • What is x-component of the total momentum of the system in kg m/s? • Assume 3 significant figures for your final answer.
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

M2 = 6.59 * -10.4

M2 = -68.536kgm/s (negative since it is moving towards the left)

The total momentum will be the sum of momentum along the x-component as shown:

M = M1+M2

M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

M = 49.4kgm/s (towards the left)

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Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

3 0
3 years ago
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
Which satellite has the greatest gravitational force with Earth?
EastWind [94]

Answer:

Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320.

Explanation:

The universal law of gravitation states that the force between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

We have to choose the satellite having greatest gravitational force with earth. In all options the distance from the earth is same i.e. 320 km. So, we have to select the satellite having maximum mass because the mass of the earth is constant.

Hence, the correct option is (D) " Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320 ".

5 0
3 years ago
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Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70
IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2

here \omega _{o}^{2}= k/m = (6.30)/(0.135) = 46.67 N/m kg

F / mA = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

\omega ^{2}= \omega _{o}^{2}\pm F/m

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π =  8.53 /6.28  or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0
3 years ago
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