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balu736 [363]
3 years ago
12

Questions -

Physics
1 answer:
irga5000 [103]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of  X =  1000 \  N and  Y  =  1400 \ N  

Explanation:

Considering the diagram,

   Taking moment about the pivot , we have

           X *  1 =  400 *  2.5

=>         X =  1000 \  N

Generally given that the bar is at equilibrium then the upward forces is equal to the down ward force

So

        Y  =  X  +  400

=>     Y  =  1000  +  400

=>     Y  =  1400 \ N    

   

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What are some modern uses for infrasound example, ultrasound can be used in the medical field to look inside people. Thanks!
lyudmila [28]
Infrasound sound is any sound that are below the sound a human can hear, elephants use it to communicate with other elephants.
5 0
3 years ago
A child with a mass of 23 kg rides a bike with a mass of 5.5 kg at a velocity of 4.5 m/s to the south. Compare the momentum of t
serg [7]

Answer:

Explanation:

Given the following data;

Mass of child = 23 kg

Mass of bike = 5.5 kg

Velocity = 4.5 m/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

To find the momentum of each of them;

I. Momentum of the child

Momentum C = mass * velocity

Momentum C = 23 * 4.5

Momentum C = 103.5 Kgm/s

II. Momentum of the bike

Momentum B = mass * velocity

Momentum B = 5.5 * 4.5

Momentum B = 24.75 Kgm/s

Hence, we can deduce from the calculations that the momentum of the child is greater than that of the bike because of the higher mass possessed by the child.

4 0
3 years ago
A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa
lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0

where:

\Delta K=+1.6\cdot 10^{-18} J is the change in kinetic energy of the particle

q=3.2\cdot 10^{-19} C is the charge of the particle

\Delta V =V_b-V_a is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V

8 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

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