experiment to calculate the spring constant of a spring.

Answer:

a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.

Explanation:

Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through.  In more rigid medium sounds velocity will be faster.

Answer: If we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.

Explanation:

A temperature at which the the liquid and gaseous phase of a substance of a substance are present in equilibrium with each other is called boiling point.

For example, the boiling point of butane is -1.5 degree Celsius.

This means that at a temperature above -1.5 degree Celsius, butane will exist is gaseous state. That is, at a temperature of 24 degree Celsius butane will evaporate.

Thus, we can conclude that if we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.

Answer:

La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.

Explanation:

Asumiendo que la varilla de cobre experimenta deformaciones muy pequeñas y que las deformaciones no longitudinales son despreciables con respecto a las deformaciones longitudinales, la deformación longitudinal de la varilla se estima mediante la siguiente fórmula:

[tex]l_{f} = l_{o}\cdot [1+\alpha \cdot (T_{f}-T_{o})][/tex] (1)

Donde:

[tex]l_{o}[/tex] - Longitud inicial de la varilla, en metros.

[tex]\alpha[/tex] - Coeficiente de dilatación, en [tex]^{\circ}C^{-1}[/tex].

[tex]T_{o}[/tex] - Temperatura inicial de la varilla, en grados Celsius.

[tex]T_{f}[/tex] - Temperatura final de la varilla, en grados Celsius.

Si sabemos que [tex]l_{o} = 1.20\,m[/tex], [tex]\alpha = 1.4\times 10^{-5}\,^{\circ}C^{-1}[/tex], [tex]T_{o} = 18\,^{\circ}C[/tex] y [tex]T_{f} = 100\,^{\circ}C[/tex], entonces la longitud final de la varilla es:

[tex]l_{f} = (1.20\,m)\cdot \left[1 + \left(1.4\times 10^{-5}\,^{\circ}C^{-1}\right)\cdot (100\,^{\circ}C-18\,^{\circ}C)\right][/tex]

[tex]l_{f} = 1.201\,m[/tex]

La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.

Answer:

Explanation:

Use the kinematic equation.

This equation can be derived from [tex]f=ma[/tex], but we can just memorize, or look them up when needed as it saves us time.

Now we can plug our measurements into each variable to solve for acceleration.

Subtract 8m/s from both sides.

Divide by 5 seconds. Left with acceleration in terms of [tex]\frac{m}{s^2}[/tex]

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

Answer:

f=ma.......m=f/a......m=20kg

Answer:

The diagram assigned B

explanation:

Check the direction of the two vectors, their resultant must be in the same direction.

Answer:

c

Explanation:

At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that

v cos(53°) = 6 m/s   ==>   v = (6 m/s) sec(53°) ≈ 9.97 m/s

The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are

x = (9.97 m/s) cos(53°) t = (6 m/s) t

y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

Find the times t for which the ball reaches a height of 3.00 m:

3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

==>   t ≈ 0.137 s   or   t ≈ 1.49 s

The second time is the one we care about, because it's the one for which the ball would be falling into the basket.

Now find the distance x traveled by the ball after this time:

x = (6 m/s) (1.49 s) ≈ 8.93 m

I hope this helps you ^-^

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

[tex]\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm[/tex]

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

[tex]\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm[/tex]

Complete Question

A light spring with force constant 3.05 N/m is compressed by 7.80 cm as it is held between a 0.400-kg block on the left and a 0.800-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push the blocks apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is 0, 0.035, and 0.397. (Let the coordinate system be positive to the right and negative to the left. Indicate the direction with the sign of your answer. Assume that the coefficient of static friction is the same as the coefficient of kinetic friction. If the block does not move, enter 0.)

(a) u = 0 heavier block m/s2 m/s2 lighter block

(b)M = 0.035 heavier block m/s2 m/s2 lighter block

Answer:

a)  [tex]A_h=0.297[/tex]

   [tex]A_l=-0.59475[/tex]

b)  [tex]a=0[/tex]

    [tex]a=-0.25175m/s^2[/tex]

Explanation:

From the question we are told that:

Force constant [tex]k=3.05N/m[/tex]

Compression Length [tex]l_c=7.80cm=0.07m[/tex]

Left Mass [tex]M_l=0.400kg[/tex]

Right Mass [tex]M_r=0.800kg[/tex]

Coefficient of kinetic friction [tex]\mu=0, 0.035, and\ 0.397.[/tex]

Therefore

Spring force is given as

 [tex]F_s=Kx[/tex]

 [tex]F_s=3.05*0.070[/tex]

 [tex]F_s=0.238N[/tex]

Generally the equation for Acceleration is mathematically given by

 [tex]A=\frac{F}{m}[/tex]

For Heavier block

 [tex]A_h=\frac{F_s}{m_r}[/tex]

 [tex]A_h=\frac{0.238N}{0.8}[/tex]

 [tex]A_h=0.297[/tex]

For Lighter blocks

 [tex]A_l=\frac{F_s}{m_r}[/tex]

 [tex]A_l=\frac{-0.238N}{0.4}[/tex]

 [tex]A_l=-0.59475[/tex]

b)

Generally the equation for Force is mathematically given by

[tex]F_s-F=ma[/tex]

For Heavier block

[tex]F>Fs[/tex]

Therefore

[tex]a=0[/tex]

For Lighter blocks

[tex]F-F_s=ma[/tex]

[tex](0.035)(0.4)(9.8)-(0.2379)=(0.4)a[/tex]

[tex]a=-0.25175m/s^2[/tex]

Answer:  

GIVEN:

v₀=0ms⁻¹

a= 8.1ms⁻²

t= 19.4s

REQUIRE:

d=?

CALCULATUION:

as we know,

d=v₀t+1/2at²

by putting values

d=0ms⁻¹×19.4s+1/2×8.1ms⁻²×(19.4s)²

d=0m+1/2×8.1ms⁻²×376.36s²

d=1/2×3048.516m

d=1524.258m

d≈1524m

Answer:

Densities increase down the group

MP and BP decrease down the group

Softness increased going down the group

Speed of reacting increases going down the group

Answer:

 ΔR = 5 s

Explanation:

The absolute uncertainty or error in an expression is

       ΔR = | [tex]\frac{dR}{dB}[/tex] | ΔB + | [tex]\frac{dR}{dA}[/tex] | ΔA

the absolute value guarantees to take the unfavorable case, that is, the maximum error.

We look for the derivatives

       [tex]\frac{dR}{dB}[/tex] = 1

       [tex]\frac{dR}{dA}[/tex] = -1

we substitute

       ΔR = 1 ΔB + 1 ΔA

of the data

       ΔB = 3 s

       ΔA = 2 s

       ΔR = 3 + 2

       ΔR = 5 s

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

[tex]a=\frac{v_f-v_i}{t}[/tex] where the change in velocity is final velocity minus initial velocity. Filling in:

[tex]3.0=\frac{v_f-(-3.0)}{6.0}[/tex] Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

[tex]3.0=\frac{v_f+3.0}{6.0}[/tex] and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

Answer:

see the explanation below

Explanation:

Momentum is a product of the mass of a particle and its velocity.

and also, momentum is a vector quantity; i.e. it has both magnitude and direction.

Now a plane in the air has both magnitude and velocity

When the plane lands the velocity will amount to zero although the mass is still very much intact

Now the mass* zero velocity= zero

Hence when a plane lands the momentum is zero

The solubility of the sample will decrease

Answer:

2000g

Explanation:

volume=mass/density

=2000/2.5

=800cm³

mass=density×volume

=800×2.5

=2000g

Answer:

Increasing

Explanation:

I Hope it Helps

Answer:

Explanation:

We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?

The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.

Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:

[tex]F_{1x}=12 cos0[/tex] and

[tex]F_{1x}=12[/tex]

[tex]F_{2x}=9cos90[/tex] and

[tex]F_{2x}=0[/tex]

[tex]F_{3x}=15 cos126.87[/tex] and

[tex]F_{3x}=-9.0[/tex]  (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).

That means that the x-component of the resultant vector, R, is 3.0

Now for the y-components:

[tex]F_{1y}=12sin0[/tex] and

[tex]F_{1y}=0[/tex]

[tex]F_{2y}=9sin90[/tex] and

[tex]F_{2y}=9[/tex]

[tex]F_{3y}=15sin126.87[/tex] and

[tex]F_{3y}=12[/tex]

That means that the y-component of the resultant vector, R, is 21.

Put them together in this way to find the resultant magnitude:

[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us

[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.

The angle, then:

[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.

To sum up: The resultant vector has a magnitude of 21 N at 82°

Answer:

cant see picture

Explanation:

Answer:

please add picture so i can help you

Explanation:

Answer:

ΔP = 1875 Pa,   P₂ = P₁ - 1875

Explanation:

Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P₁ -P₂ = ½ ρ (v₂² - v₁²)

suppose the fluid is water

          P₁ - P₂ = ½ 1000 (2² - 0.5²)

         ΔP = 1875 Pa

this is the pressure difference between the two sections

the pressure in the narrowest section is

           P₂ = P₁ - 1875

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

Answer:

The time required is 0.45 s.

Explanation:

Height, h = 1 m

initial velocity, u = m/s

Let the time is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\1 = 0 +0.5 \times 9.8 \times t^2\\\\t = 0.45 s[/tex]