student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.

Answer:

If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

Explanation:

Answer:

answer is C on edge 2021

Explanation:

Answer:

There are three temperature scales in use today, Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water, the interval between the two being divided into 180 parts

Answer: I have don't understand this question

Explanation:

Answer:100m west

Explanation:

Answer:

20.0V

SANA MAKATULONG

The voltage drop across the 10.0-ohm resistor would be 20.0 volt.

Resistance is the obstruction of electrons in an electrically conducting material.

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

For calculating equivalent resistance in series combination.

Re = R1  + R2 + R3

For the given problem the total resistance of the circuit would be as all the three resistors are connected in the series combination.

Re= 10+20+30

Re=60 ohm

As given in the figure all the resistance are connected in the series combination therefore the current flowing through them would be the same.

For the given problem we have to design and construct a circuit that has two resistors connected in series.

By using Ohms law

V=IR

120 = 60×I

I = 2 ampere

Given that we have a 120 V battery, that will produce a current of 2 Ampere

By using Ohm's law we can calculate the voltage drop across a 10-ohm resistor

V=IR

  =2×10

  = 20 volt

Thus, the voltage drop across the 10.0-ohm resistor comes out to be 20.0 volt.

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Answer:

a

Explanation:

all the other ones take years and hard work but learning japanese is pretty easy and you dont need to work to hard

The answer is A, B, D.

Answer:

The answer is A, B, D.

Explanation:

ap3x verified

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

Answer:

451 milliliters equals 15.25 fluid ounces

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{c*b}{a}[/tex]

The direct rule of three is the rule applied in this case where there is a change of units.

In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?

[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]

fluid ounces= 15.25

451 milliliters equals 15.25 fluid ounces

Answer:

e. 55.0 m

Explanation:

Given;

diameter of the aluminum wire, d = 0.865 mm

radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m

resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m

resistance of the wire, R = 2.48 Ω

The resistance of a wire is given by;

[tex]R = \frac{\rho \ L}{A} \\\\[/tex]

where;

L is length of the wire

A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²

Substitute the givens and solve for L,

[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]

Therefore, the length of the wire is 55.0 m

Answer:

5,000

Explanation:

Vf = Vi + a * t

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

Answer:half-life (T1/2) of this isotope =

Explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material  at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 =  e−k(4.50)

In 0.3767  =   -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is  related to the rate constant by

T₁/₂=ln  2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln  2/0.2159

T₁/₂=3.12 days

Answer:

The value is [tex]I = 0.48 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The coordinate of the knee is  [tex]knee = (1.2,2.7)[/tex]

    The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]

    The mass of the shank is  [tex]m = 3.2 \ kg[/tex]

 Generally the length between the knee and the ankle is mathematically represented as

      [tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]

=>  [tex]D =0.67 \ m[/tex]

Generally the  moment of inertia of the shank about the proximal end point is mathematically represented as

         [tex]I= m * \frac{D^2}{3}[/tex]

=>       [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]

=>       [tex]I = 0.48 \ kg m^2[/tex]

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Weight of the object is force applied due to gravity, So force = mass.gravity .

on the  moon gravity is 1/6 of that on the earth.

so weight or force = 80*(9.8/6)

                               = 130 Newtons

The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.

To learn more about force refer to the link:

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Answer:

a. Magnification = 6.1

b. The image formed is virtual, and on the same side of the lens as the object.

c. Image size = 119.8 squared millimetres

Explanation:

Magnification = [tex]\frac{Image distance}{Object distance}[/tex]

But, focal length, f = 15.0 cm, and object distance, u = 12.4 cm. Let the image distance be represented by v.

a. Applying the lens formula, we have;

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{15}[/tex] = [tex]\frac{1}{12.4}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{15}[/tex] - [tex]\frac{1}{12.4}[/tex]

  = -[tex]\frac{13}{930}[/tex]

v = -75.1538

The image distance, v = -75.2 cm

Magnification = [tex]\frac{75.2}{12.4}[/tex]

                       = 6.0645

Magnification = 6.1

b. The image formed is virtual, and on the same side of the lens as the object.

c. Given that diameter of mole = 5.00 mm.

Its radius = [tex]\frac{diameter}{2}[/tex] = [tex]\frac{5.0}{2}[/tex]

              = 2.5 mm

Thus, the area of the mole would be;

A = [tex]\pi[/tex][tex]r^{2}[/tex]

   = [tex]\frac{22}{7}[/tex] x [tex](2.5)^{2}[/tex]

   = 19.643

A = 19.64 square millimetres.

Thus, the size of the image can be determined by;

Magnification = [tex]\frac{Image size}{Object size}[/tex]

Image size = Magnification x object size

                  = 6.1 x 19.64

                  = 119.804

The size of the image is 119.8 squared millimetres.

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer:

A Magnesia

Explanation:

The word magnetism comes from the word Magnesia, which is the name for a region in Asia Minor, where fragments of Fe3O4 ore (magnetite) were found, which attracts other metal objects.

Magnetism is a physical phenomenon by which we describe the attractive or repulsive force between materials.

This phenomenon has been known for thousands of years.

Answer:

-20°C

Explanation:

The specific heat capacity of ice using the cgs system is 0.5cal/g°C

The enthalpy change is calculated as follows

ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.

10cal = 2g×0.5cal/g°C×∅

∅=10cal/(2g×0.5cal/g°C)

∅=10°C

Final temperature= -30°C+ 10°C= -20°C

Answer:

The final temperature of the bullets is 327.3 ºC.

Explanation:

Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:

[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)

Where:

[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.

[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.

[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.

[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.

By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)

Where:

[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.

[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.

[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.

[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.

[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.

Now we clear the final temperature of the bullets:

[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]

[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]

[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:

[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]

[tex]T = 852.534\,^{\circ}C[/tex]

Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)

[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:

[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]

The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:

[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)

Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.

If we know that  [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:

[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]

[tex]U_{max} = 557.375\,J[/tex]

Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.

Answer:

The reason is similar to the reason why it is difficult to roll an object on a surface with a positive incline than rolling it on the ground

The more the path becomes vertical, the more force we have to apply to oppose the force of gravity

but when we are moving horizontally, we don't have to move against the gravity and hence, it is less difficult than going vertically upwards

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➡ 150hrs.

[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

Answer:

The constant force to be exerted on the rope is 221.55 N

Explanation:

Given;

mass of the merry, m = 235 kg

radius, r = 1.5 m

number of revolution per second, = 0.4 rev/s

time of motion, t= 2.00 s

The angular acceleration is given by;

[tex]\alpha = \frac{0.4 \ rev}{s} *\frac{2\pi \ rad}{rev} *\frac{1}{2.0\ s} = 1.257 \ rad/s^2[/tex]

Torque is given by;

τ = F.r

Also torque in uniform solid disk is given by;

τ = ¹/₂mr²α

Thus, equating the two;

F.r = ¹/₂mr²α

F =  ¹/₂mrα

F = ¹/₂(235)(1.5)(1.257)

F = 221.55 N

Therefore, the constant force to be exerted on the rope is 221.55 N

Answer:

depends on how talll the building is but lets say its 100 ft tall 12MPH

Explanation:

Answer:

around 9.81m/s i think

Explanation:

Answer:

distance travelled = 3000m

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]