What is the acceleration when a force of 2 N is applied to a ball that has a mass of 0.60 kg?

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

Answer:

The energy is transferred to chemical and heat energy.

Explanation:

If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. This is the proof of the answer i Hope this helps :)

Answer:

Explanation:

.....

Answer:

MGH=energy

3500*9.8*h=90000

h=90000/34300

h=2.62m

Answer:

no kinetic energy

hope this helps! :-D

Explanation:

the monk is not moving

Answer:

26.9444m/s

pls brainliest

Answer:

c

Explanation:

Answer:−4.05

Explanation:

Answer:

63°

that's my answer

but then I am sorry if I'm wrong

Explanation:

90-27 = 63°

Answer:

       y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Explanation:

Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation

             λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]

Let's use conservation of energy for speed

starting point

             Em₀ = U = e V

final point

             Em_f = K = ½ m v²

             Em₀ = Em_f

             eV = ½ m v²

             v =[tex]\sqrt{\frac{2eV}{m} }[/tex]

we substitute

             λ=  [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]

the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum

            a sin θ = m λ

first zero occurs at m = 1, also these experiments are performed at very small angles

            sin θ = θ

            θ = λ / a

This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain

           θ = 1.22 λ / D

where D is the diameter of the opening

we substitute

          θ = [tex]\frac{1.22}{D}[/tex]   \sqrt{ \frac{h^2 m}{2eV}}  

this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians

            θ = y / L

when substituting

where y is the minimum distance that can be resolved for this acceleration voltage

            y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Answer:

Explanation:

Angle of incidence at the longer face = 45⁰ ( see the figure in the attached file )

For total reflection from longer face

i = C where C is critical angle .

And relation between critical angle and refractive index is as follows .

μ = 1 / sinC

= 1 / sin45

= √2

= 1.414 .

Answer:

The heavier the object and the higher it is above the ground, the more gravitational potential energy it holds. Gravitational potential energy increases as weight and height increases. Potential energy is energy that is stored in an object or substance.

Answer:

Explanation:

Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration

T₁ sin52 + T₂ sin40 = 3 x 9.8

.788 T₁ + .643 T₂ = 29.4

The horizontal component of tension will add up to zero because the decoration piece is at rest .

T₁ cos52 + T₂ cos40 = 0

.615 T₁ -  .766 T₂ = 0

T₁ =  1.24 T₂

Substituting this value of T₁ in earlier equation , we have

.788 x 1.24 T₂ + .643 T₂ = 29.4

1.62 T₂ = 29.4

T₂ = 18.15 N

T₁ = 1.24 x 18.15 = 22.51 N .

maybe a spectrograph ?

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

Answer:

the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

Explanation:

Given;

mass of the particle, m = 1.43 g = 0.00143 kg

electric field experienced by the particle, E = 700 N/C

The force experienced by the particle is calculated as;

F = mg = EQ

Where;

Q is the magnitude of the charge

[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]

The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.

Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.

The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.

Given information-

The mass of the particle is 1.43 g or 0.00143 kg.

The magnitude of the downward-directed electric field is 700 N/C.

The magnitude of the free-fall acceleration is 9.80 meter per second squared.

The electric field is defined as the electric force per unit charge. It can be given as,

[tex]E=\dfrac{F}{q}[/tex]

Rewrite the equation for the charge,

[tex]q=\dfrac{F}{E}[/tex]

Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,

[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]

Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.

Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

Learn more about the electric charge here;

https://brainly.com/question/14372859

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

[tex]F=Eq[/tex]

[tex]ma=Eq[/tex]

For the proton

[tex]m_{p}a_{p}=Eq_{p}[/tex]

[tex]a_{p}=\frac{Eq_{p}}{m_{p}}[/tex]

[tex]a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}[/tex]

[tex]a_{p}=6.19*10^{10}\: m/s^{2}[/tex]

For the electron

[tex]m_{e}a_{e}=Eq_{e}[/tex]

[tex]a_{e}=\frac{Eq_{e}}{m_{e}}[/tex]

[tex]a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}[/tex]  

[tex]a_{e}=1.14*10^{14}\: m/s^{2}[/tex]

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

[tex]x_{p}+x_{e}=0.0426[/tex]

Both of them have an initial speed equal to zero. So we have:

[tex]\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426[/tex]

[tex]t^{2}(a_{p}+a_{e})=2*0.0426[/tex]

[tex]t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}[/tex]

[tex]t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}[/tex]

[tex]t=2.73*10^{-8}\: s[/tex]    

With this time we can find the distance from the positive plate (x(p)).

[tex]x_{p}=\frac{1}{2}a_{p}t^{2}[/tex]

[tex]x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}[/tex]

[tex]x_{p}=0.0023\: cm[/tex]

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

Answer:

a

Explanation:

i just took the test

Answer:

  E = V / d

Explanation:

In a charged capacitor an electric field is established that goes from the positive to the negative plate, this field is constant,

the potential difference is

           D = E d

in this case they do not give the difference in potential V and the distance between the plates d

           E = V / d

Answer: 30 Years Old

Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.

Answer:

30.4 s

Explanation:

A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g

a = 4 x 9.8 = 39.2 m /s²

initial velocity u = 0

Final velocity = 3.60 times speed of sound

= 3.6 x 331 = 1191.6 m /s

v = u + at

Putting the values

1191.6 = 0 + 39.2 t

t = 30.4 s .

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

Answer:

a) d= 1393 m

b) θ= 21º S of E.

Explanation:

a)

       [tex]d = \sqrt{(1300m)^{2} + (500m)^{2} } = 1393 m (1)[/tex]

b)

       [tex]tg \theta = \frac{500m}{1300m} = 0.385 (2)[/tex]

       ⇒ θ= tg⁻¹ (0.385) = 21º S of E.

HERE IS YOUR ANSWER!

Answer:

case A) tau_net = -243.36 N m, case B)    tau_net = 783.36 N / m,      tau_net = -63.36 N m,  case C)  tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

         Σ τ  = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

         tau_net = m g x - F x2

          tau_nett = -28.8 9.8 1.5 + 180 1

         tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

 the force is up

          tau_nett = -28.8 9.8 1.5 - 180 2

          tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

        tau_nett = -28.8 9.8 1.5 + 180 2

         tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

 tau_nett = -28.8 9.8 1.5 - 180 3

 tau _net = - 963.36 N m

3 counterclockwise

       tau_nett = -28.8 9.8 1.5 + 180 3

       tau_net = 116.64 Nm

the sitam rotated counterclockwise

Answer:

Explanation:

Average velocity = Total displacement / total time

Average speed = total distance covered / total time

a )

For the entire trip from your front door to the bench

Total displacement = 55 - 35 = 20 m  [ first displacement is positive and second displacement is negative , because second displacement is in opposite direction ]

Total displacement = 20 m

Total time = 30 + 36 = 66 s

Average velocity = 20 / 66

= .303 m / s

b )

For the entire trip from your front door to the bench

Total distance covered  = 55 + 35 = 90  m

Total time = 30 + 36 = 66 s

Average speed  = 90 / 66

= 1.36  m / s

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³