When scientists want to determine the actual age of a fossil, they use_____________

Answer:

We know that the gravitational force F between two masses P and Q, that are separated by a distance R is:

[tex]F = G*\frac{P*Q}{R^2}[/tex]

Where G is the gravitational constant.

a) Mass P is doubled, then we have 2*P instead of P, the new force is:

[tex]F' = G*\frac{(2*P)*Q}{R^2} = 2*(G*\frac{P*Q}{R^2} ) = 2*F[/tex]

b) Now R is doubled, then instead of R, we have 2*R:

[tex]F' = G*\frac{P*Q}{(2*R)^2} = G*\frac{P*Q}{4*R^2} = G*\frac{P*Q}{R^2}*(1/4) = F/4[/tex]

c) Now we replace P by 2*P, and Q by 3*Q

[tex]F' = G*\frac{(2*P)*(3*Q)}{R^2} = 2*3*(G*\frac{P*Q}{R^2} ) = 6*F[/tex]

d) The entire mass of the system is increased by a factor of 4, then both of the individual masses are increased by a factor of 4.

Then we need to replace P by 4*P, and Q by 4*Q.

[tex]F' = G*\frac{(4*P)*(4*Q)}{R^2} = 4*4(G*\frac{P*Q}{R^2} ) = 16*F[/tex]

e) Now we replace R by R/2.

[tex]F' = G*\frac{P*Q}{(R/2)^2} = G*\frac{P*Q}{R^2/4}= 4*G\frac{P*Q}{R^2} = 4*F[/tex]

Answer:

the number of grains in the ball is 274,848

Explanation:

Given that;

diameter = 0.5 mm

so radius r = 0.25 mm

first we determine the volume of the ball using the following equation;

V = 4/3×πr³

we substitute

V = 4/3×π(0.25)³

V =  0.06544 mm³

Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³

so;

Number of grains N = 0.06544 × 4,200,000

N = 274,848 grains

Therefore, the number of grains in the ball is 274,848

Answer:

Gravitation Potential Energy and Elastic Potential Energy

Explanation:

PE=abbreviation for potential energy

PE is a mechanical energy like KE is. There are two types of PE in high school physics (do not talk about other PE's like chemical PE).

Elastic PE is like the stored energy for example in a pole vault competition, the Elastic PE is stored in the pole before the vaulter goes over the desired height. (Energy stored in ELASTIC object ONLY) (commonly on the 11th grade regents springs are used to represent elastic PE)

Gravitational PE can be calculated through the formula PE=mgh although some people substitute the PE with a U. (ENERGY STORED IN OBJECT'S HEIGHT)

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W =  (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

Answer: The difficulty comes in the modulation, and in the fact that X-rays can't be transmitted via wires or fiber.

Explanation: Mark me as brainlisti

Answer:

Δp = 620 N.s

Explanation:

       [tex]F = \frac{\Delta p}{\Delta t} (1)[/tex]

       for m= constant, it reduces to F= m*a.

       [tex]\Delta p = F* \Delta t = 62 N* 10.0 s = 620 N*s (2)[/tex]

Answer:

98

Explanation:

Answer:

friction, air resistance

Explanation:

Answer:

It begins to increase

Explanation:

As the surfaces move slowly and then increase their relative speeds, the coefficient of kinetic friction begins to increase until it reaches it terminal value in which, it can no longer increase.

Answer:

Explanation:

For a string of length L clamped down on both ends , for fundamental mode of vibration

2 . λ / 4 = L

= λ / 2 = L

λ = 2 L

For other modes , the formula is

n . λ/2 = L where n is an integer .

λ = 2L / n

if n = 6

λ = 2L / 6

= L/3

So wavelength = L/3 is possible .

Answer: Location the data and observations collected in your lab guide

Explanation:

Answer: 20

Explanation: cause

Explanation:

When two or more atoms link up, they create a molecule. e.g, A molecule of water is made of two atoms of hydrogen (H) and one atom of oxygen (O).

Answer: C. A stove burner that glows red when it is turned on

Explanation:

I did the quiz

Answer:

0m/s²

Explanation:

The acceleration of the given body whose motion has been described will be zero.

This is because the body moves with a constant velocity.

Acceleration is defined as the rate of change of velocity with time.

Since in this problem, the velocity is not changing per time, then the acceleration would be zero.

A body moving steadily implies that such body is not changing its motion by the virtue of its velocity and even direction.

Answer:

Psychology traces its roots back through recorded history to India, China, the Middle East, and Europe. Buddha and Confucius focused on the powers and origin of ideas. ... The ideas of Bacon and Locke contributed to the development of modern empiricism.

Explanation:

Answer:

3k mph

Explanation:

don't take my answer it is wrong

Answer:

The answer is :(B) 1.9 N

please make my answer brainlist

Answer:

Explanation:

Let the velocity of firing be u at angle θ

At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ

So , vtop = v cosθ

At height h/2

vertical component of velocity v₂

v₂² = (usinθ)² - 2 g . h/2

v₂² = u²sin²θ - gh

horizontal component of velocity at height h/2 = u cosθ

velocity at height h / 2

= √ ( u²sin²θ - gh + u² cos²θ)

Given

√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop

u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ

u²sin²θ - gh =  3 u² cos²θ

At height h , vertical component of velocity is zero

0 = u²sin²θ - 2gh

gh = u²sin²θ / 2

u²sin²θ - u²sin²θ / 2  =  3 u² cos²θ

u²sin²θ / 2  = 3 u² cos²θ

Tan²θ = 6

Tanθ = 2.45

θ = 68⁰ .  

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

The amount of work that must be done to stop a hoop of mass 110 kg moving with s speed of 0.220 m/s is 2.662 J

Work can be defined as the ability or the capacity to perform work. Work is also the measure of energy of a body. The S.I unit of work is Joules (J)

Formular for kinetic energy of the hoop is

EK = mv²/2.................... Equation 1

Where EK = Kinetic energy of the hoop, m = mass of the hoop, v = velocity of the hoop.

Given: m = 110kg, v = 0.220 m/s.

Substitute these values into equation 1

EK = 110(0.22)²/2

Ek = 2.662 J.

Therefore, The amount of work that must be done on the hoop to stop it is

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Answer:

the  ratio of r₂/r₁ is 1.37.

Explanation:

Given;

magnitude of electric field strength at a distance r₁, E₁ = 300 N/C

magnitude of electric field strength at a distance r₂, E₂ = 160 N/C

Electric field strength is given as;

[tex]E = \frac{kQ}{r^2} \\\\E_1r_1^2 = E_2r_2 ^2\\\\\frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \\\\\frac{E_1}{E_2} =( \frac{r_2}{r_1})^2\\\\\sqrt{\frac{E_1}{E_2} } = \frac{r_2}{r_1}\\\\\frac{r_2}{r_1} = \sqrt{\frac{300}{160} }\\\\\frac{r_2}{r_1} =1.37[/tex]

Therefore, the  ratio of r₂/r₁ is 1.37.

Answer:

146.27 m

Explanation:

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

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Value of d =?

Next, we shall determine the time taken for tourist and the bear to get to the car.

For the tourist:

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Time (tₜ) =?

vₜ = dₜ / tₜ

4.65 = d/tₜ

Cross multiply

4.65 × tₜ = d

Divide both side by 4.65

tₜ = d / 4.65

For the bear:

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Time (t₆) =?

v₆ = d₆ / t₆

5.68 = (32.4 + d) / t₆

Cross multiply

5.68 × t₆ = (32.4 + d)

Divide both side by 5.68

t₆ = (32.4 + d) / 5.68

NOTE: Both the tourist and the bear have the same time. Thus, to obtain the value of d, we shall equate both the time taken for the tourist and the bear together. This is illustrated below:

Time taken by tourist (tₜ) = time taken by the bear (t₆)

tₜ = d / 4.65

t₆ = (32.4 + d) / 5.68

tₜ = t₆

d / 4.65 = (32.4 + d) / 5.68

Cross multiply

d × 5.68 = 4.65 (32.4 + d)

5.68d = 150.66 + 4.65d

Collect like terms

5.68d – 4.65d = 150.66

1.03d = 150.66

Divide both side by 1.03

d = 150.66 / 1.03

d = 146.27 m

Thus, the maximum value of d is 146.27 m

The strength of the electric field is 1.10×10^4 N/C.

The data is sent to us.

r = d/2 = 0.1/2 = 0.05 m, d = 10 cm = 0.1 m, and z = 17 cm = 0.17 m

Q = 40 nC = 40×10^-9 C

A) in the middle of the two circles

Since the electric fields created by the two rings are directed in opposite directions, there is no electric strength.

B) midway down the left ring

Electric strength is equal to the sum of the electric fields created by the left and right rings.

Electric strength is equal to E leftring + E rightring, which equals 0 plus kQz/(r2 + z2).

Electric strength = k×Q×z/(r^2 + z^2)^

Electric strength = 9×10^9×40×10^-9×0.17/(0.05^2 + 0.17^2)

Electric strength= 1.10×10^4 N/C

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Answer: The more massive one will have larger kinetic energy.

Explanation:

We know that when we drop an object, the acceleration of the object will be equal to the gravitational acceleration.

a(t) = -9.8m/s^2

And to get the velocity, we need to integrate over time, to get:

v(t) = (-9.8m/s^)*t + v0

Where v0 is the initial speed of the object.

You can see that the mass of the object does not affect the velocity of it.

Then when we drop two marbles of different masses from the same height, we know that the final velocity of them will be equal.

Now, we also know that the kinetic energy can be written as:

K = (m/2)*v^2

where m is the mass, and v is the velocity.

Then the kinetic energy of the marble with less mass can be written as:

k = (m/2)*v^2

And the kinetic energy for the more massive one is:

K = (M/2)*v^2

And we know that both of them have the same velocity, and M is larger than m, then we can conclude that the marble with larger mass will have larger kinetic energy.

Answer:

B. contain hydrogen

C. clouds of gas and dust

E. needed to create a star

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

Some of the examples of stars are; Vega, Sun (closest to planet Earth), Antares, Betelgeus, Canopus, etc.

Stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He). The chronological order in which the formation of a star occur are;

1. Gravity pulls gas and dust together to form dense cores.

2. A protostar forms as mass increases.

3. Nuclear fusion begins under high pressure.

Scientists have been able to understand and discover that, gravity pulled materials (low-density cloud of interstellar gas and dust known as a nebula) together forming the planetary bodies in our solar system.

A dark nebula can be defined as an interstellar cloud that is so dense as a result of high concentration of gas and dust and as such it obscures the visible wavelengths of light from stars behind it, thus appearing completely opaque (dark patch) in front of a bright emission nebula or in regions having plenty stars.

The characteristics of a nebulae are;

I. It contain hydrogen.

II. Clouds of gas and dust

III. It is needed to create a star.

Answer:

For two waves of equal amplitude interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave. For 100 waves of the same amplitude interfering constructively, the resulting amplitude is 100 times larger than the amplitude of an individual wave.

Answer:

The sign of the velocity depends on the coordinate system chosen to define the position. A positive velocity simply means that the object is moving in the positive direction, as defined by the coordinate system, while a negative velocity means the object is traveling in the other direction.

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Answer:

2.61 atm

Ley de Boyle

Explanation:

[tex]P_1[/tex] = Presión inicial = 0.96 atm

[tex]P_2[/tex] = Presión final

[tex]V_1[/tex] = Volumen inicial = 95 mL

[tex]V_2[/tex] = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

[tex]\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}[/tex]

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

Answer:

See the explanation below,

Explanation:

Momentum is defined as the product of mass by Velocity. In this way, we will replace the following equation in each of the cases.

[tex]P=m*v[/tex]

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

A)

[tex]P = 10000*0\\P=0[/tex]

B)

[tex]P=100*5\\P=500[kg*m/s][/tex]

C)

[tex]P=1200*15\\P=18000[kg*m/s][/tex]

D)

[tex]P=15*1000\\P=15000[kg*m/s][/tex]

From higher to lower momentum

C,D,B,A

Answer:

The magnitude of the maximum net force on the object during the motion is 16 newtons.

Explanation:

The position of the mass, measured in meters, within the mass-spring system is:

[tex]x(t) = A\cdot \cos \left(\omega\cdot t \right)[/tex] (1)

Where:

[tex]A[/tex] - Amplitude, measured in meters.

[tex]t[/tex] - Time, measured in seconds.

[tex]\omega[/tex] - Angular frequency, measured in radians per second.

The acceleration function ([tex]\ddot {x} (t)[/tex]), measured in meters per square second, is obtained by deriving twice (1) in time:

[tex]\dot {x} (t) = -\omega\cdot A\cdot \sin (\omega\cdot t)[/tex]

[tex]\ddot {x} (t) = -\omega^{2}\cdot A\cdot \cos (\omega\cdot t)[/tex] (2)

And the maximum acceleration ([tex]a_{max}[/tex]), measured in meters per square second, experimented by the mass is:

[tex]a_{max} = \omega^{2}\cdot A[/tex] (3)

And the maximum net force ([tex]F[/tex]), measured in newtons, is:

[tex]F = m\cdot a_{max}[/tex] (4)

If we know that [tex]\omega = 2\,\frac{rad}{s}[/tex], [tex]A = 0.80\,m[/tex] and [tex]m = 5\,kg[/tex], then the maximum net force on the object during the motion is:

[tex]a_{max} = \left(2\,\frac{rad}{s} \right)^{2}\cdot (0.80\,m)[/tex]

[tex]a_{max} = 3.2\,\frac{m}{s^{2}}[/tex]

[tex]F = (5\,kg)\cdot \left(3.2\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 16\,N[/tex]

The magnitude of the maximum net force on the object during the motion is 16 newtons.

The magnitude of the maximum net force on the object during the motion will be 16 newtons.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The given data in the problem is;

m is the mass of  5.0 of object kg

x (t ) = A cos(ωt ),

A is the amplitude=  0.80 m

ω is the frequency= 2.0 s-1 .

F is the  magnitude of the maximum net force=?

The position of the mass is given by the equation;

[tex]x(t)= A cos( \omega t)[/tex]]

[tex]\em \dot x= - \omega A sin(\omega t) \\\\ \dot \dot x = - \omega ^2 A cos (\omega t)[/tex]

The maximum acceleration is found as;

[tex]\rm a_{max} = \omega ^2 A \\\\ \rm a_{max} = (2) ^2 \times 0.80 \\\\ \rm a_{max} =3.2 \m/sec^2[/tex]

The value of net force is found as;

[tex]\rm F= m a_{max} \\\\ \rm F= 5 \times 3.2 \\\\\ \rm F=16\ N \\[/tex]

Hence the magnitude of the maximum net force on the object during the motion will be 16 newtons.

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