The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?

Answer:

the one in which the fluid has the lowest density

Answer:

Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.

Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).

Answer:

The race car will crash into the outer wall

Explanation:

max fr = μsN = 1.1 mg = 11 m

mv2/R = m(100)2/(400) = 25 m > fr

Answer:

Explanation:

20 n is an unknown amount

If that is supposed to be 20 N(ewtons)

then W = Fd = 20(15) = 300 J

Answer: it will be 300 newton meters

Explanation:

Answer:

The chemical structure of the molecule is responsible for each of these characteristics. The chemical structure is comprised of the bonding angle, the kind of bonds, the size of the molecule, and the interactions that occur among the molecules. Even little changes in the chemical structure of a molecule may have a significant impact on the characteristics of the substance.

Explanation:

Hope it helps:)

Answer:

a = 25 m/s2

Explanation:

A = f/m

A = Speed/Acceleration

F =‘Force

M = Mass

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the  force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is  3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

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Answer:

200 kilocalories

Explanation:

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

[tex]x=\frac{m_{steam}}{m_{total}}[/tex]

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

[tex]x=\frac{2kg}{10kg} =0.20[/tex]

Now, at this temperature and pressure, the volume of a saturated vapor is  2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]

This means that the volume of the container will be:

[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]

Learn more:

The force of friction needed to overcome to move the box is 29.4N

According to Newton's second law;

[tex]\sum F_x = ma_x\\[/tex]

Taking the sum of force along the plane;

[tex]F_m -F_f = ma\\F_m -F_f = 0\\F_m=F_f = \mu R[/tex]

This shows that the moving force is equal to the frictional force

Given that

[tex]\mu = 0.6\\R = mg = 49N[/tex]

Get the frictional force;

Since

[tex]F_f = \mu R\\F_f = 0.6 \times 49\\F_f = 29.4N[/tex]

Hence the force of friction needed to overcome to move the box is 29.4N

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Explanation:

The buoyant force [tex]F_B[/tex] is defined as

[tex]F_B = \rho_wgV[/tex]

where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:

[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]

Plugging in the values, we get

[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]

Recall that the weight of an object in terms of its density and volume is given by

[tex]W = \rho gV[/tex]

Using the value for the volume above, we can solve for the density of the object as follows:

[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]

[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]

Answer:

39 volts

Explanation:

Use the equation [tex]P=VI[/tex]

[tex]78=V(2)[/tex]

[tex]V=39[/tex]

The Celcius Scale is called an absolute Temperature scale,and it's zero point called absolute zero

_______________

Answer:

3

Explanation:

state of matter are solid

liquid and

gases

Answer:

3

Explanation:

state of a matter are solid liquid and gas

Answer:

0.80 kN

Explanation:

Hope you understood it

Answer:

0

Explanation:

0 is the answer

Explanation:

In this chapter, we will study the important concepts of kinetic energy and the

closely related concept of work and power.

A- Kinetic Energy

Kinetic energy is a physical quantity, which is associated with the moving objects

and defined as:

K = ½ mv2

If the body is stationary (v=0), its kinetic energy is zero. The SI unit of kinetic

energy is kg.m2

/s2

or Joule (J), where 1 J=1 kg.m2

/s2

. Kinetic energy is a scalar

quantity.

B- Work

The work is defined as the ability to perform a force along a certain displacement.

There are different types of work as follows:

1- Work done by a constant force

The work done by the constant force F is given by the scalar product of the force F

and the displacement d.

WF = F.d = Fd cosθ

where θ is the angle between the force and displacement. The above equation means

that the work is the product of the displacement magnitude by component of the

force parallel to the displacement. Therefore, work is a scalar quantity (only

magnitude, no direction) and can be positive, negative, or zero. The SI unit of work

is (N.m) or joule (J) where 1 N.m = 1 J.

Special cases and remarks:

• If the angle between the force and displacement is zero (parallel), the work is

WF = F d (maximum work)

Answer:

V2 = (V1 - u) / (1 - V1 u / c^2)

V1 = speed of ship in observer frame = .99 c    to right

u = speed of frame 2 = -.99 c   to left relative to observer

V2 = speed of V1 relative to V2

V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c

V2 = 1.98 / (1 + .99^2) c = .99995 c

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 20 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]

The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]

Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of  m·r² increases, the value of [tex]\omega _f[/tex] decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

https://brainly.com/question/15188750

Answer:

Explanation:

½(75)v² = ½(0.008)390²

        v² = (0.008)390²/75

        v² = 16.224

         v = 4.027...

         v = 4.0 m/s

Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]

[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]

Plug this value into the equation for kinetic energy:

[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]

Part C:

The potential energy of a spring is given as:

[tex]U = \frac{1}{2}kx^2[/tex]

Find 'k' using the derived quantity above:

[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]

Now, calculate potential energy:

[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]

Answer:

A 5.00- kg block is placed on top of a 10.0 -kg block (Fig. P5.68). A horizontal force of 45.0 N is applied to the 10.0-kg block, and the 5.00- kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks.

(b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block.

The horizontal acceleration of the block is 5 m/s².

To calculate the horizontal acceleration on the block, we use the formula below.

Formula:

Where:

Make "a" the subject of the equation.

Substitute these values into equation 2

Substitute these values into equation 2

Hence the horizontal acceleration of the block is 5 m/s².

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Hello!

To solve, we can begin by using the kinematic equation:

[tex]d = v_it + \frac{1}{2}at^2[/tex]

Where:

vi = initial velocity (m/s)

t = time (s)

a = acceleration (in this case, due to gravity. g = 9.8 m/s²)

Since the object falls from rest, the initial velocity is 0 m/s.

[tex]d = \frac{1}{2}at^2[/tex]

Plug in the given values:

[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]

Answer:

This happens when the mass of one of the objects is very large and does not move.

Explanation:

Answer:

Explanation:

At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.

0.75g = ω²R

R = 0.75g/ω²

R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²

R = 8.283006...

R = 8.28 m