A student is investigating the affect of different salts on melting points. Four patches of ice of equal
size are roped off and a
different type of salt is poured on each, one receives table salt (NaCl), one
receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth
receives inert sand instead. Each patch receivęs an equal amount of salt or sand. The student
measures the volume of ice remaining and subtracts it from the original volume of ice to see how
much melted away. What is a control variable in this experiment?
A. The size of the ice patches.
B. The type of salt applied to the ice.
C. The amount of ice that melted.
D. None of these.

Answer:

0.4

Explanation:

A concave lens is a diverging lens, so it will always have a negative focal length. Image distance is always negative for a concave lens because it forms virtual images.

From the lens formula;

1/f = 1/u+ 1/v

- 1/2 = 1/3 - 1/v

1/v = 1/3 + 1/2

v= 6/5

v= 1.2 cm

Magnification = image distance/object distance

Magnification = 1.2cm/3cm

Magnification = 0.4

Mendeleev wrote down atomic mass

Answer:

A. Buoyancy.

hope it helps

Answer:

the third one T-W

Explanation:

the direction of the Tension and weight are opposite

Answer:

(n) alluvial fan sandbar

Explanation:

Answer:

2.82 m/s²

Explanation:

[tex]v = u + at \\ 24 = 0 + a(8.5) \\ a = 2.82 \: ms {}^{ - 2} [/tex]

The acceleration of a motorcycle is 2.82 m/s^2.

Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates.

Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.

Given that:

Initial velocity of the motor cycle: u = 0 m/s.

Final velocity of the motorcycle = 24 meter/second.

Time taken to reach this velocity = 8.5 second.

Hence, acceleration of the motor cycle = change in velocity/time interval

= ( final velocity - initial velocity)/time interval

= ( 24 m/s  - 0 m/s)/8.5 s

= 2.82 m/s^2.

Its acceleration is 2.82 m/s^2.

Learn more about acceleration here:

brainly.com/question/12550364

#SPJ2

Answer:

The package should be dropped 244.7 meters before the target

Explanation:

We need to find how long the package will take to hit the ground then solve for position.

y=0

-4.9t^2=-3000

t= sqrt (3000/4.9)

t= 24.74 seconds.

x= 247.4 meters.

Answer:

The answer is C.

Explanation:

Answer:

t = 3.44 s

Explanation:

We are given;

Fired from rest, and so; u = 0 m/s

Final speed; v = 170 m/s

Height above flat ground; y_o = 58 m

Height at starting point; y = 0 m

Thus, from Newton's equation of motion, we have;

y - y_o = ut - ½gt²

(since it's motion is against gravity)

Plugging in the relevant values, we have;

0 - 58 = 0 - (½ × 9.8 × t²)

-58 = -4.9t²

t² = 58/4.9

t = √(58/4.9)

t = 3.44 s

Answer:

what if I do and b then someone else c I don't have enough time pls

Answer:

Yes

Explanation:

Many scientists believe that the universe is expanding at an increasing rate. Scientists cannot explain how the universe is expanding at an increasing rate because by law of conservation of energy, there is only a finite amount of energy. Thus, scientists have called this new source of energy dark matter and it is all around the universe helping shape and form the universe.

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 300 mA = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

[tex]C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF[/tex]

Capacitance of capacitor 2,

[tex]C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF[/tex]

Capacitance of capacitor 3,

[tex]C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF[/tex]

(b) The equivalent capacitance in series combination is :

[tex]\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF[/tex]

Hence, this is the required solution.

Answer:

-standard 1 kg : Kilogram (kg), basic unit of mass in the metric system.

-standard 1 meter: The standard metre is the length of the path travelled by light in vaccum during a time interval of 1/299792458 of a second.

- standard 1 second : The second (abbreviation, s or sec) is the Standard International ( SI ) unit of time.

Explanation:

HOPE IT HELPS YOU !!

Answer:

Attenuation

Step By step Explanation:

The decrease in the intensity of light over distance, such as with increasing depth in water, is known as ______.

It is called attenuation.

The decraese in the amplitude of the signals is called attenuation.

Explanation:

Hydraulic Pressure-Control, On-Off Deluge Valve

FP-400Y-5DC

The BERMAD model 400Y-5DC is an elastomeric, hydraulic line pressure operated deluge valve, designed specifically for advanced fire protection systems and the latest industry standards. The 400Y-5DC is activated by a hydraulically operated relay valve, through which opening and closing of the valve can be controlled either with a remote hydraulic command or with a wet pilot line with closed fusible plugs. An integral pressure reducing pilot valve ensures a precise, stable, pre-set downstream water pressure. The optional valve position indicator can include a limit switch suitable for Fire & Gas monitoring systems. The 400Y-5DC is ideal for systems that combine a remote wet pilot line with a high pressure water supply.

Answer:

C

Explanation:

Pigments reflect some wavelength of visible light and absorb

Answer:

2 Hz.

Explanation:

Frequency is simply defined as the number of appearances of a periodic event occurring per time. It is usually measured in cycles/second.

Now, in this question, we are told that there are 2 cycles for each second.

Thus, we can say that the frequency is 2 cycles/1 s = 2 Hz.

Answer:

In coin card experiment smooth card is used so that the card can slide easily from glass

Answer:

acceleration = - 150 m/s^2

distance = 9 miles.

Explanation:

initial speed, u = 60 mph

time, t = 12minutes = 0.2 hour

final speed, v = 30 mph

Let the acceleration is a and the distance is s.

By the first equation of motion

v = u + at

30 = 60 + a x 0.2

a = - 150 m/s^2

Let the distance is s.

Use third equation of motion is

[tex]v^2 = u^2 + 2 a s \\\\30^2 = 60^2 + 2 \times 150\times s\\\\s = 9 miles[/tex]

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

The plane has a centripetal acceleration a of

a = v ²/r

where v is the plane's tangential speed and r is the radius of the circle. By Newton's second law,

F = mv ²/r

Solve for the mass m :

m = Fr/v ² = (3000 N) (18.3 m) / (55.0 m/s)² ≈ 18.1 kg

Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction.

Answer:

3420.39 N

Explanation:

Applying,

Fd = 1/2(mv²-mu²)................. Equation 1

Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.

make F the subject of the equation

F = (mv²-mu²)/2d............... Equation 2

From the question,

Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m

Substitute these values into equation 2

F = [(890×0²)-(890×1.4²)]/(2×0.255)

F = -1744.4/0.51

F = -3420.39 N

The negative sign denotes that the force in opposite direction to the motion of the car.

Answer:

the car’s final displacement is 60 m

Explanation:

Given;

initail velocity of the car, u = 0

acceleration of the car, a = 0.8 m/s²

time of motion, t = 10 s

The first displacement of the car:

[tex]x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m[/tex]

The second displacement of the car;

acceleration, a = 0.4 m/s²,   time of motion, t = 10 s

[tex]x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m[/tex]

The final displacement of the car;

x = x₁  +  x₂

x = 40 m  +  20 m

x = 60 m

Therefore, the car’s final displacement is 60 m

Answer:

Frequency = 24 × 10⁸ Hz

Explanation:

Given the following data;

Speed = 3 × 10⁸ m/s

Wavelength = 0.125 meters

To find the frequency of the electromagnetic wave;

Mathematically, the speed of a wave is given by the formula;

Speed = Wavelength × frequency

Substituting into the formula, we have;

3 × 10⁸ = 0.125 × frequency

Frequency = (3 × 10⁸)/0.125

Frequency = 24 × 10⁸ Hz

Answer:

Explanation:

Assuming the squirrel is jumping off the ground, here's what we know but don't really know...

v₀ = 4.0 at 50.0°

So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:

[tex]v_{0y}=4.0sin(50.0)[/tex] which gives us that the upward velocity is

v₀ = 3.1 m/s

Moving on here's what we also know:

a = -9.8 m/s/s and

v = 0

Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:

v = v₀ + at and filling in:

0 = 3.1 - 9.8t and

-3.1 = -9.8t so

t = .32 seconds.

Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:

Δx = [tex]3.1(.32)+\frac{1}{2}(-9.8)(.32)^2[/tex] and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:

Δx = .99 - .50 so

Δx = .49 meters

Answer: prevent excitation light from reaching the detector

Explanation:

A fluorometer refers to the device that's used in the measurement of parameters that are of visible spectrum fluorescence. They also prevent excitation light from reaching the detector.

These parameters are used in the identification of the amount and presence of molecules in a medium.

Answer:

im pretty sure it should be 50.0

Answer:

S = 1/2 g t^2      where t is the time to fall 1.2 m

t = (2 S / g)^1/2 = (2 * 1.2 / 9.8) = .495 s

Sy = Vy T = 3 m/sec  * .495  sec = 1.48 m     distance from edge of table

(Rotational speed has no effect since table is smooth)