If mass is the amount of MATTER present in the
object, will it be different on Earth and the moon?
No
Yes

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:

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

Explanation:

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

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:

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

Answer:

825m

Explanation:

u=20m/s

a=0.5m/(s)^2

s = ut + 1/2a(t)^2

s = 20(30) + 1/2(0.5)(30)^2

s = 600 + 225

s = 825m

Answer:

as we know that

S=ut+1/2(at*t)

S=20*30+1/2(0.5*30*30)

S=600+225

S=825

Answer:

theory

Explanation:

took the quiz

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:

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

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:

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

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:

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:

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

"Energy" is the ability to do work.

"Work" is the process of using energy.

Answer:

second column

Explanation:

Answer:

Trial 3 is the answer.

Explanation:

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:

1) They are electrically neutral

2) They have slightly more weight than protons

Explanation:

The given atomic particle found in the nucleus has the following characteristics;

The location of the particle = The nucleus

The (numbers of the) particle does not contribute to (change) the atomic number of the element

The particles found within the nucleus of an atom are; Neutrons and protons

The particle within the nucleus that determines the atomic number = The number of protons

Therefore, the particle referenced in the question is the neutrons

The two characteristics of the neutron are;

1) The neutrons are neutral, electrically

2) Neutrons have slightly more weight than protons

3) Neutrons are magnetic

4) Neutrons are very small

5) Neutrons consist of three quarks; One 'Up', and two 'Down' quarks

Therefore, two characteristics of the particle are;

1) They are electrically neutral and 2) They are slightly heavier than protons.

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:

5.33ms-¹

Explanation:

that is the procedure above

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:

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:

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:

you have written the definition so what are you asking

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

[tex]E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\[/tex]

Now equate the forces,

[tex]T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree[/tex]

Answer:

below

Explanation:

that is the procedure above

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:

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Explanation:

..........

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