A boy travels 12km east wards to a point B and then 5km southwards to another point C. Calculate the difference between the magnitude of the displacement of the boy and the distance travelled by him

"Energy" is the ability to do work.

"Work" is the process of using energy.

Answer:

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Explanation:

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Answer:

second column

Explanation:

Answer:

Trial 3 is the answer.

Explanation:

Answer:

[tex]\triangle \phi=461.5rad[/tex]

Explanation:

From the question we are told that:

Silt width [tex]w=0.15=>0.1510^{-3}[/tex]

Wavelength [tex]\lambda=462nm=462*10^{-9}[/tex]

Angle [tex]\theta=26.9[/tex]

Generally the equation for Phase difference is mathematically given by

[tex]\triangle \phi=\frac{2 \pi}{\lambda}(\frac{wsin\theta }{2})[/tex]

[tex]\triangle \phi=\frac{2 \pi}{462*10^{-9}}(\frac{0.1510^{-3}*sin 26.9 }{2})[/tex]

[tex]\triangle \phi=461.5rad[/tex]

[tex]\triangle \phi=146.89\pi[/tex]

Answer:

Time, t = 2 seconds

Explanation:

Given the following data;

Mass, m = 50 kg

Initial velocity, u = 0 m/s (since it's starting from rest).

Final velocity, v = 8 m/s

Force, F = 200 N

To find the time, we would use the following formula;

[tex] F = \frac {m(v - u)}{t} [/tex]

Making time, t the subject of formula, we have;

[tex] t = \frac {m(v - u)}{F} [/tex]

Substituting into the formula, we have;

[tex] t = \frac {50(8 - 0)}{200} [/tex]

[tex] t = \frac {50*(8)}{200} [/tex]

[tex] t = \frac {400}{200} [/tex]

Time, t = 2 seconds

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

Answer:

Etc....

Explanation:

#keep learning!!

Explanation:

POTENTIAL energy is the energy that is stored in an object due to its position relative to some zero position.

Answer:

2 more neutrons

Explanation:

To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:

For carbon–14:

Mass number = 14

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

14 = 6 + Neutron

Collect like terms

14 – 6 = Neutron

8 = Neutron

Neutron number = 8

For carbon–12:

Mass number = 12

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

12 = 6 + Neutron

Collect like terms

12 – 6 = Neutron

6 = Neutron

Neutron number = 6

SUMMARY:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Finally, we shall determine the difference in the neutron number. This can be obtained as follow:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Difference =?

Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)

Difference = 8 – 6

Difference = 2

Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)

Answer:

[tex]{ \bf{power \: { \tt{(watts)}} = \frac{workdone \: { \tt{(joules)}}}{time \: { \tt{(seconds)}}} \: }}[/tex]

Answer:

theory

Explanation:

took the quiz

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

[tex]F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}[/tex] I'm going to do the math on the top and then on the bottom and divide at the end.

[tex]F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}[/tex] and now when I divide I will express my answer to the correct number of sig dig's:

[tex]Fg=[/tex] 6.45 × 10¹⁶ N

Answer:

Explanation:

The equation for acceleration is

[tex]a=\frac{v_f-v_0}{t}[/tex] which is the final velocity minus the initial velocity divided by the time. I first need to put the units all in the same terms. We have the velocity given as km/hr, but the time is given in seconds and that's not gonna work. I will change the velocity to km/sec:

[tex]90\frac{km}{hr}*\frac{1hr}{3600s}=.025\frac{km}{s}[/tex] That's the value we will use for the final velocity of this car in the equation for acceleration:

[tex]a=\frac{.025-0}{10}=.0025\frac{km}{s^2}[/tex]

The second part of this problem asks how far the car travels in this 10 seconds. We just determined that the car can travel .025 km in 1 second, so in 10 seconds the car travels 10(.025) = .25 km

Answer:

Explanation:

This is a Law of Momentum Conservation problem where, in particular, our problem looks like this:

[tex][m_dv_d+m_uv_u]_b=[m_dv_d+m_uv_u]_a[/tex] in other words, the momentum before they collide has to be equal to the momentum after they collide. Knowing that the dragon is initially at rest:

[1000(0) + 50(600)] = [1000(40)m + 50v] and

0 + 30,000 = 40,000 + 50v and

-10,000 = 50v so

v = -200 m/s or

200 m/s in the direction opposite to its initial direction

Answer:

5.33ms-¹

Explanation:

that is the procedure above

Answer:

The gravitational potential energy (gpe) possessed by an object or body is directly proportional to the height of the object or body.

Explanation:

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Generally, the gravitational potential energy (gpe) of an object or body is directly proportional to the height of the object or body. Thus, the gravitational potential energy of a body increases as the height of the body increases.

In conclusion, an object with a higher height would have a higher gravitational potential energy.

Answer: hi your question is incomplete  attached below is the complete question

answer :

magnitude of acceleration :  | a | = g = 9.81 m/s^2

direction : a = - g j

Explanation:

Neglecting Air resistance

magnitude of acceleration :

| a | = g = 9.81 m/s^2

Direction of acceleration

a = - g j  ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )

Answer:

Option 1 & 2

Explanation:

The area of mathematics known as Boolean algebra concerns with operations on logical quantities with binary variables. To express truths, Boolean features are transformed as binary numbers: 1 = true and 0 = false. Boolean algebra concerns with logistical processes, whereas fundamental algebra deals with machine based.

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

Explanation:

Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time

Answer:

a. 15,000J

b. .76m

Explanation:

KE = (1/2)m*v²

KE = .5*1000kg*30m/s

KE = 15000J

PE = m*g*h

7500J = 1000kg*9.81m/s²*h

7500J = 9810*h

h = .76m

Answer: 2.43 s

Explanation:

Initial velocity is [tex]u=15.2\ m/s[/tex]

Final velocity [tex]v=23\ m/s[/tex]

Average acceleration is [tex]a_{avg}=3.2\ m/s[/tex]

Average acceleration is change in velocity in the given amount of time

[tex]\therefore a_{avg}=\dfrac{v-u}{t}\\\\\Rightarrow 3.2=\dfrac{23-15.2}{t}\\\\\Rightarrow t=\dfrac{7.8}{3.2}\\\\\Rightarrow t=2.43\ s[/tex]

Thus, 2.43 s is required to acquire that average acceleration with 23 m/s velocity .

Answer:

Explanation:

There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.

Going with the physics version first, here's what we know:

a = -9.8 m/s/s

v₀ = 3.75 m/s

t = ??

That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:

v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:

v = v₀ +at and filling in:

0 = 3.75 + (-9.8)t and

-3.75 = -9.8t so

t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:

2(.38) = .76 seconds. Now onto the calculus way:

The position function is

[tex]s(t)=-4.9t^2+3.75t[/tex] The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:

v(t) = -9.8t + 3.75 and

0 = -9.8t + 3.75 and

-3.75 = -9.8t so

t = ,38 and multiply that by 2 to find the time the whole trip took:

2(.38) = .76 seconds.

There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.

Answer:

you have written the definition so what are you asking

Answer:

779.87 N/m²

Explanation:

Pressure = Force (F) / Area (A)

Recall :

Force = ma ; mass * acceleration due to gravity

a = 9.8 m/s² ; mass = 0.4kg

Force, F = 0.4 * 9.8 = 3.92 N

Area = πr² ; r = Radius = 4cm = 4 / 100 = 0.04 m

Area = π0.04² = 0.0050265 m²

Hence,

Pressure = 3.92 / 0.0050265

Pressure = 779.86670 N/m²

Pressure = 779.87 N/m²

Answer:

d=1/2 (a)(t^2)

D = distance

A = acceleration

T = time

acceleration due to gravity is 32 ft/second

so, d=1/2 (32)(5^2)

d=16(25)

d=400

Explanation:

Answer:

400 ft.

Explanation:

D= 1/2 gt^2

=1/2(-32 ft/sec^2)(5 sec^2)

= -(1/2)(32)(25) ft

D= -400 ft, down

Answer:

a= 2.5m/s²

Explanation:

U=0

V=90km/hr

T= 10s

Convert 90km/hr to m/s

1km= 1000m

1hr= 3600s

(60×60)

therefore, 90km/hr = 90000/3600

90km/hr= 25m/s

From Newton First Equation,

V=U + AT

25=0+ A(10)

25= 10A

25/10 =10A/10

A= 2.5m/s²

Answer:

To explain what happens with the ball we must remember the Law of Conservation of Energy.

This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.

Then,

At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.

When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.

At the middle of the hill, both energies have the same values.

At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.

Correct me if I'm wrong...

hope it helps...

Answer:

Part A

a)  F = -16x + 4,  b)  x = 0.25 m, c) STABLE

Explanation:

Part A

a) Potential energy and force are related

          F = [tex]- \frac{dU}{dx}[/tex]- dU / dx

          F = - (8 2x -4)

          F = -16x + 4

b) The object is in equilibrium when the forces are zero

          0 = -16x + 4

          x = 4/16

          x = 0.25 m

c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.

In this case there is only one equilibrium point

by changing the position a bit

           x ’= x + Δx

we substitute

          F ’= - 16 x’ + 4

          F ’= - 16 (x + Δx) + 4

          F ’= (-16x +4) - 16 Δx

at equilibrium position F = 0

          F ’= 0 - 16 Δx

we can see that the body returns to the equilibrium position, therefore it is STABLE

PART B

This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved

initial instant. Before the shock

        p₀ = m v

final instant. After the crash

        p_f = (m + M) v_f

We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic

          p₀ = pf

          mv = mv ’+ M v_f

in the case of the elastic collision, the kinetic energy is conserved

        K₀ = K_f

        ½ m v² = ½ m v’² + ½ M v_f²

we write the system of equations

        mv = mv ’+ M v_f (1)

         m (v² -v'²) = M v_f ²

         m (v - v ’) = M v_f

         m (v-v ’) (v + v’) = M v_f

        v + v ’= v_f

we substitute in equation 1 and solve

         v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]

         v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]

the mechanical energy of the neutron is

  initial

          Em₀ = K = ½ m v²

final moment

          Em_f = K + U = ½ m v_f ² + U

U is the energy lost in the collision

total energy is conserved

          Em₀ = Em_f

          ½ m v² = ½ m v_f ² + U

         U = ½ m (v² -v_f ²)

         U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex]  v)² ]

       U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]

       U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]

       U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]

Let's do the same calculations for the nucleus

initial     Em₀ = 0

final        Em_f = K + U = ½ M v_f ² + U

            Em₀ = Em_f

            0 = K + U

            U = -K

            U = - ½ M v_f ²

            U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²

            U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]  

We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.

b) the fraction of energy lost

          f = U / Em₀

          f = 4 m M / m + M        

c) let's calculate the fraction of energy lost in a collision

          m = 1.67 10⁻²⁷ kg

          M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg

         f = 4 1.6 20 / (1.6+ 20)    10⁻²⁷

         f = 5.92 10⁻²⁷ J

the energy of a fast neutron is greater than 1 eV

         Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J

Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo

         #_collisions = 0.95 Eo / f

         #_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷

         #_collisions = 2.7 10⁷ collisions