Một vật chuyển động thẳng biến đổi đều đi hết quãng đường AB trong thời gian 6 giây. Vận tốc của vật khi đi qua điểm A là 5m/s, khi đi qua điểm B là 15m/s. Tìm chiều dài của quãng đường AB. A. 50 m B. 80 m C. 60 m D. 70 m →→→

Answer:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge

Answer: Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge. Thus, an object's charge can be exactly 0 e, or exactly 1 e, −1 e, 2 e, etc., but not, say, 12 e, or −3.8 e

Explanation:

Answer:

d)

Explanation:

the only one that makes sense

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

The weight of the satellite at a distance of one Earth radius above the surface of Earth is 40 N.

The given parameters:

The gravitational force on the satellite is calculated by applying Newton's law of universal gravitation is calculated as;

[tex]F = \frac{Gm_1m_2}{R^2} \\\\F_1 R_1^2 = F_2R_2^2\\\\F_2 = \frac{F_1 R_1^2 }{R_2^2} \\\\F_2 =\frac{200 \times R_1^2}{(2R_1)^2} \\\\F_2 = \frac{200 \times R_1^2 }{4R_1} \\\\F_2 = 40 \ N[/tex]

Thus, the weight of the satellite at a distance of one Earth radius above the surface of Earth is 40 N.

Learn more about Newton's law of universal gravitation here: https://brainly.com/question/9373839

no net force axting on it

Answer:

I think none of above

Explanation:

hioe it's right answer

Assume R is measured in meters (m) and M in kilograms (kg). Then

R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²

so t ² is indeed proportional to R ²/(GM).

Answer:

Resistance is a measure of the opposition to current flow in an electrical circuit.All materials resist current flow to some degree. They fall into one of two broad categories: Conductors: Materials that offer very little resistance where electrons can move easily.

Answer:

Depends: Is it tennis or volleyball?

Explanation:

If it's volleyball, it's C, the other teams point.

If it's tennis, it's A, you get a re-do.

Answer:

G is universally constant

Explanation:

The Gravitational constant G, is a proportionality constant for all objects in the universe with a value of ; G = 6.67*10^-11 m³kg^-1s^-2.

According to the the Newton's Law of Universal Gravitation ; Every object in the universe attracts one another with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.

F α m1m2 α 1/r²

F α m1m2/r²

F = Gm1m2 / r²

m1 and m2 are the masses ; r = distance between them

G = Gravitational constant.

Answer:

Explanation:

Begin by remembering that at the stone's max height, the final velocity there is 0. Being in possession of that information along with the fact that the pull of gravity due to acceleration is -9.8 m/s/s, we can use the equation

[tex]v^2=v_0^2+2a[/tex]Δx where v is the final velocity, v₀ is the initial velocity, a is the pull of gravity, and Δx is the displacement. Filling in:

[tex]0^2=v_0^2+2(-9.8)(5.0)[/tex]

0 = v₀² - 98 so

98 = v₀² and

v = 9.9 m/s Now we can put that into the KE equation.

[tex]10=\frac{1}{2}m(9.8)[/tex] and

[tex]m=\frac{2(10)}{98}[/tex] so

m = .20 kg

Answer:

m

Explanation:

.

that's according to ma observation

Ohm's law states that :

the terms used are :

let's solve for electric current :

[tex]\mathfrak{good\: \: luck \: \: for \: \: your \: \: assignment}[/tex]

Answer:

dₒ/dᵢ = 0.32

Explanation:

From the question given above, the following data were obtained:

Object height (Hₒ) = 4 mm

Image height (Hᵢ) = 1.25 cm

Ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] =?

Next, we shall convert 4 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

4 mm = 4 mm × 1 cm / 10 mm

4 mm = 0.4 cm

Next, we shall determine the magnification of the lens. This can be obtained as follow:

Object height (Hₒ) = 4 mm

Image height (Hᵢ) = 1.25 cm

Magnification (M ) =?

M = Hᵢ/Hₒ

M = 1.25 / 0.4

M = 3.125

Finally, we shall determine the ratio of object distance (dₒ) to image distance (dᵢ). This can be obtained as follow:

Magnification (M) = 3.125

Ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] =?

Magnification (M) = image distance (dᵢ) / object distance (dₒ)

M = dᵢ/dₒ

Invert

1/M = dₒ/dᵢ

1/3.125 = dₒ/dᵢ

dₒ/dᵢ = 0.32

Therefore, the ratio of object distance (dₒ) to image distance (dᵢ) [dₒ/dᵢ] is 0.32

Answer:

Workdone = 19600 Nm

Explanation:

Given the following data;

Mass = 40 kg

Distance = 50 meters

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the work done;

First of all, we would determine the force being exerted by the construction worker.

Force = mass * acceleration due to gravity

Force = 40 * 9.8

Force = 392 Newton

Next, we would determine the work done;

Workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

[tex] Workdone = 392 * 50 [/tex]

Workdone = 19600 Nm

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

[tex]f=\frac{v}{\lambda}[/tex] and filling in:

[tex]2.5=\frac{v}{2.0}[/tex] so

v = 2.5(2.0) and

v = 5.0 m/s

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

Answer:

i) v = 1.47 cm/s

ii) distance = 219.8 cm

iii) displacement = 28cm

Explanation:

Remember that:

The motion of the tip of the second hand is a circular motion.

For something that rotates with an angular frequency ω, and a radius R, the velocity is given by:

v = ω*R.

i) We know that the second hand of a clock does a complete rotation each 60 seconds.

Then the period is:

T = 60s

And the frequency is the inverse of the period, so:

f = 1/T

f = (1/60s)

And the angular frequency is 2*pi times the normal frequency, thus:

ω = 2*pi*f

ω = 2*pi*(1/60s)  = (2*pi/60s)

And the radius will be 14 cm, the velocity of the ant is:

v = (2*pi/60s)*14cm

if we replace pi by 3.14 we get:

v =  (2*3.14/60s)*14cm = 1.47 cm/s

ii) The distance covered by the ant in 150 seconds:

Remember that the period of the clock is T = 60s

so in 150 seconds we have:

150s/60s = 2.5 revolutions.

Then the total distance covered is 2.5 times the perimeter of a circle of radius R = 14cm, this is:

distance = (2.5)*2*pi*14cm

              = (2.5)*2*3.14*14cm = 219.8 cm

iii) We want to know the displacement, this is, the difference between the final position and the initial position.

In 150 seconds, the ant does 2.5 revolutions.

So the ant will end in the opposite side of the circle where she started (if the ant started when the second hand was at the "3", then the final position is when the second hand is at the "9").

So the displacement will be equal to twice the radius, or the diameter of the circle.

if the radius is 14cm, the diameter is:

2*14cm = 28cm

the displacement is 28cm

Answer:

a) the distances are zero, Both 1st & 2nd condition

c)  the torques are equal but of the opposite sign, 2nd condition of equilibrium

Explanation:

The equilibrium conditions are

1 translational

        ∑ F = 0

2 rotational

        ∑ τ = Σ (F_i x r_i) = 0

They tell us that external torque is zero.

Therefore we have two various possibilities

a) the distances are zero, in this case we have a pure translation movement

    for this situation the two equilibrium relations are fulfilled

b) the forces are zero, there is no movement

    It does not make sense to use the equilibrium relations since there are no forces

c) the torques are equal but of the opposite sign, the forces are on the opposite side of the body.

    In this case the 2 equilibrium relation is fulfilled, but not the first one that the force has the same direction

Answer:

3.44×10-27 joules.

Explanation:

E = h * c / λ = h * f,

E is the energy of a photon

h is the Planck constant

c is the speed of light

λ is the wavelength

Answer:

Yes

Explanation:

If there were to be a supernova of a star or a planet/meter that was directed at earth, then we would know in advance and make a plan to stop it

Well since the gravitational pull is so strong, nothing can escape from it which means we haven't been able to get close enough to really know everything about them. There's no way of knowing what truly goes on inside of them since there is no way out.

Answer:

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

Explanation:

Answer: on jupiter you would weigh the most  and on mars you would weigh the least

Explanation:

2nd egg experienced more impulse

Hope this helps! :)

Answer:

E = 3.8 kJ

Explanation:

Given that,

The mass of the object, m = 10 g = 0.01 kg

The heat of fusion of  aluminum is 380 kJ/kg

We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:

E = mL

So,

E = 0.01 × 380

E = 3.8 kJ

So, the energy required to melt the mass is equal 3.8 kJ.