A small, solid sphere of mass 0.2 kg and radius 40 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 1.5 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 40 cm. What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop?

Answer:

this is how I know it though my answer is not in the choices

Explanation:

M=80g

V=37cm-12 cm

15cm

Density=mass/volume

80g/15cm3

5.33g/cm3

Explanation:

Given

mass of the rock is 10 kg

Force requires to hold the rock is equal to its weight

Weight is given by the product of mass and acceleration due to gravity

Weight on the earth surface

[tex]\Rightarrow W_e=10\times 9.8\\\Rightarrow W_e=98\ N[/tex]

Weight on the moon surface

[tex]\Rightarrow W_m=1.6\times 10\\\Rightarrow W_m=16\ N[/tex]

So, the force holding the rock on earth is approximately 6 times the force on the moon.

The cross-sectional area of the cylindrical copper wire is  1.7 x 10⁻⁹ m².

Resistivity is defined as the amount of electrical resistance of a conductor, per unit cross-sectional area and unit length.

Here,

Resistance of the wire, R = 10 Ω

Length of the wire, l = 1 m

Resistivity of the wire, ρ = 1.7 x 10⁻⁸ Ωm

The expression for the resistivity of a conductor is given as,

ρ = RA/l

where A is the cross-sectional area of the conductor.

Therefore, the area of cross-section of the copper wire,

A = ρl/R

A = 1.7 x 10⁻⁸ x 1/10

A = 1.7 x 10⁻⁹ m²

Hence,

The cross-sectional area of the cylindrical copper wire is  1.7 x 10⁻⁹ m².

To learn more about resistivity, click:
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Answer:

The dotted line connecting two points is known as the distance traveled.

Explanation:

The distance traveled is a physical quantity that refers to the total displacement of a body. This distance is usually shown as a dotted line that connects two points in an image. In general, we can say that the distance traveled is calculated by adding all the distances that an element makes when leaving one point and reaching another point, which is its final position.

An example of this can be seen in the image shown below, where the red line represents the displacement of a body that left point A and arrived at point B. The blue dotted line represents the distance traveled and refers to the total displacement of that body, showing all the distances it covered while walking from point A to point B.

Answer:

D. Nuclear Fission and alpha decay

Answer:

a = 0 m/s²

Explanation:

From A to B the jeep travels in a straight line with a constant speed of 10 m/s. Acceleration is defined as the rate of change in velocity. In the current scenario, the velocity of the jeep remains constant. The direction of the jeep does not change as it travels in the straight path. Also, the magnitude of the velocity is constant at 10 m/s.

Therefore, there is no change in the velocity of the jeep from A to B. Hence, the will be no acceleration from A to B.

a = 0 m/s²

Answer:

The correct answer is D C A

Answer:

the rate of change of velocity per unit of time.

Answer:

accelerate is rate of chang of velocity

Answer:

Ants position at point A = -7 cm

Ants position at point B = 38 cm

Ants position at point C = 26 cm

Ants position at point D = -7 cm

Explanation:

In this question, we are dealing with displacement which is the change in position of an object.

Now, we are told the journey began from A to B and then from C to D.

Now, AB = 31 cm and AO = 7cm

But along AB, we have the origin O.

Since A is on the left hand side of the origin, it means it is negative. Thus, position A = -7cm

Then position B = 31 - (-7) = 31 + 7 = 38 cm

Since BC = 12cm, then position C = 38 - 12 = 26 cm

Position D is same as position A = -7cm

Answer:

1568 J

Explanation:

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

Mass (m) = 80 Kg

Distance (d) = 5 m

Coefficient of kinetic friction (μ) = 0.4

Workdone (Wd) =?

Next, we shall determine the normal reaction. This can be obtained as follow:

Mass (m) = 80 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) =?

N = mg

N = 80 × 9.8

N = 784 N

Next, we shall determine force of friction. This can be obtained as follow:

Coefficient of kinetic friction (μ) = 0.4

Normal reaction (N) = 784 N

Force of friction (F) =?

F = μN

F = 0.4 × 784

F = 313.6 N

Finally, we shall determine the work done. This can be obtained as follow:

Distance (d) = 5 m

Force of friction (F) = 313.6 N

Workdone (Wd) =?

Wd = F × d

Wd = 313.6 × 5

Wd = 1568 J

Thus, the workdone is 1568 J

Answer:

λ = 266 nm

Explanation:

In this case, we need to determine first the separation of the second fringe from the central maximum. We can determine that with the following expression:

d sinθ = mλ   (1)

However, as the slits are very narrow, we can assume that sin θ ≈ θ and so θ = x/l

Replacing this in (1) we have:

d(x/l) = mλ

and solving for x:

x = mλl / d    (2)

Where:

x: separation of the 2nd fringe from the central maximum

m: order of the fringe

λ: wavelength of light

l: distance of the screen

d: distance between the slits

All the units must be in meters (m), so we can convert the units first or during the resolution. In this case, we'll do it in the resolution. Replacing the given data, we have:

x = 2 * (520 nm * 1m/10⁻⁹ nm) * 1.6 m / (0.66 mm * 1m/1000 mm)

x = 0.00252 m or just 2.52 mm

With this value, we can compute or determine the separation of the 2nd order fringe of the unknown light in the central maximum:

x₂ = 2.52 - 1.23 = 1.29 mm or 0.00129 m

Now, using (2) we can solve for λ:

λ = dx / ml   (3)

Replacing we have:

λ = (0.00129 * 0.0066) / (2 * 1.6)

Hope this helps

Answer:

F₁ = 499.61 N , this is the force that Bubba support

Explanation:

The trunk is in equilibrium with the two forces applied by man, let's use the equilibrium relation

let's set a reference frame at the extreme left and assume that the counterclockwise rotations are positive

Let's write the expression for the translational equilibrium

subscript 1 is for Bubba's mass and subscript 2 for his partner

               F₁ + F₂ -W = 0

                F₁ + F₂ = W

the expression for rotational equilibrium

               ∑ τ = 0

               F₁  2.2 + F₂  (6.2-0.9) - W  6.2/2 = 0

               2.2 F1 + 5.3 F2 = 3.1 W

let's write our system of equations

             F₁ + F₂ = W

             2.2 F₁ + 5.3 F₂ = 3.1 W

we solve for F₁ in the first equation and substitute in the second

            F₁ = W-F₂

            2.2 (W- F₂) + 5.3 F₂ = 3.1 W

            F₂ ( -2.2 +5.3) = W (3.1 - 2.2)

            F₂ = 704  0.9 / 3.1            

            F₂ = 204.39 N

This is the force that the partner supports

we look for F1

            F₁ = W-F₂

            F₁ = 704 - 204.39

             F₁ = 499.61 N

This is the force that Bubba support

Answer:

[tex]0.01\ \text{A}[/tex]

[tex]0.0015\ \text{C}[/tex]

[tex]0.0608\ \text{s}[/tex]

Explanation:

[tex]V_0[/tex] = Voltage = 1.5 V

[tex]C[/tex] = Capacitance = [tex]1000\ \mu\text{F}[/tex]

[tex]R[/tex] = Resistance = [tex]150\ \Omega[/tex]

Current is given by

[tex]I=\dfrac{V_0}{R}\\\Rightarrow I=\dfrac{1.5}{150}\\\Rightarrow I=0.01\ \text{A}[/tex]

Current flowing in the resistor is [tex]0.01\ \text{A}[/tex].

Charge is given by

[tex]Q=CV\\\Rightarrow Q=1000\times 10^{-6}\times 1.5\\\Rightarrow Q=0.0015\ \text{C}[/tex]

The charge on the capacitor is [tex]0.0015\ \text{C}[/tex].

Voltage is given by

[tex]V=V_0e^{-\dfrac{t}{RC}}\\\Rightarrow t=-RC\ln\dfrac{V}{V_0}\\\Rightarrow t=-150\times 1000\times 10^{-6}\times\ln\dfrac{1}{1.5}\\\Rightarrow t=0.0608\ \text{s}[/tex]

Time taken to reach 1 V is [tex]0.0608\ \text{s}[/tex].

Answer:

A stormy ocean, having the most kinetic energy means it moves more and faster

Answer: 0.553

Explanation:

Answer:

(a) 0.014 A

(b) 2 V

Explanation:

(a) Applying

V = IR'...................... Equation 1

Where V = Voltge, I = current, R = total resistance.

make I the subject of the equation

I = V/R'................... Equation 2

From the question,

Given: V = 12 V, R' = (122+232+500) ohms (The resistance are connected in series) = 854 ohms

Substitute these values into equation 2

I = 12/854

I = 0.014 A

(b)  Applying

V' = V(R₁)/(R₁+R₂+R₃)...................... Equation 3

Where V' = Voltage across the 100 ohms resistor.

From the question,

V = 12V, R₁ = 100 ohm, R₂ = 200 ohm, R₃ = 300 ohm.

Substitute these values into equation 3

V' = (12×100)/(100+200+300)

V' = 1200/600

V' = 2 V

Answer:

v = 4.233  m/s

Explanation:

By applying the rate of boiling from [tex]Q= mL_v[/tex];

the rate of the boiling can be described as:

[tex]\mathcal{P} = \dfrac{Q}{\Delta t} \\ \\ \mathcal{P} = \dfrac{mL_v}{\Delta t}[/tex]

The mode of the steam (water vapor) as an ideal gas can be illustrated by formula:

[tex]P_oV_o = nRT[/tex] --- (1)

where;

n = number of moles;

[tex]n = \dfrac{mass (m)}{Molar mass (M)}[/tex]

Then; equation (1) can be rewritten as:

[tex]P_oV_o = (\dfrac{m}{M}) RT \\ \\ \dfrac{P_oV}{\Delta T} = \dfrac{m}{\Delta t} ( \dfrac{RT}{M})[/tex]

∴

[tex]\dfrac{m}{\Delta t} = \dfrac{\mathcal{P}}{L_v}[/tex]

Then:

[tex]P_o \times A \times v= \dfrac{\mathcal{P}}{L_v}\Big ( \dfrac{RT}{M }\Big)[/tex]

making (v) the subject of the formula:

[tex]v= \Big ( \dfrac{\mathcal{P} RT}{M\times L_v \times P_o \times A }\Big)[/tex]

Given that:

[tex]\mathcal{P}[/tex] = 0.90  kW = 900 W

R(rate constant) = 8.314 J/mol.K

Temperature at 100° C = 373K

molar mass= 18.015 g/mol ≅ 0.0180 kg/mol

Latent heat of vaporisation [tex]L_v[/tex] = 2.26 × 10⁶ J/kg

Atmospheric pressure [tex]P_o = 1.013 \times 10^6 \ N/m^2[/tex]

Cross sectional area A =1.60 cm² = 1.60 × 10⁻⁴ m²

[tex]v= \Big ( \dfrac{900 W (8.314 \ J/mol.K)(373)}{0.0180 \ kg/mol) (2.26 \times 10^6 \ J/kg) (1.013 \times 10^5 \ N/m^2)(1.60 \times 10^{-4} \ m^2)}\Big)[/tex]

v = 4.233  m/s

Answer:

Option A. 160 cm tall and 100 cm in front of Amy

Explanation:

To know the the correct answer to the question, we must recognise that the image formed by plane mirror is:

I. Same size as the object.

II. Same distance as the object from the mirror.

III. Laterally inverted.

IV. Virtual

With the above information, we obtained the answer to the question as follow:

Object height = 160 m

Thus,

Image height = 160 m

Distance of object from mirror = 50 cm

Therefore,

Distance of image from the mirror = 50 cm

Distance between the object and image = 50 + 50 = 100 cm

Thus, option A gives the correct answer to the question because the image of Amy is 160 cm tall and 100 cm in front of her.

Answer:

I think its d matter accreted from a companion star unstable ignites on the surface of a white dwarf.

Answer:

c

Explanation:

units c mark me brainless k

Answer:

Explanation:

(a) Electron Emission (Beta- decay):

When an unstable nucleus decays by the emission of Beta- particle, its charge number ‘Z’ increases by 1 but, its mass number ‘A’ remains unchanged. The transformation is represented by the equation:  

[tex]_zX^A\ -------->\ _{Z+1}Y^A\ +\ _{-1}e^0[/tex]

It is called ‘Negative Beta Decay’. It is more common than alpha decay.

Example:

[tex]_6C^{14}\ -------->\ _{7}N^{14}\ +\ _{-1}e^0[/tex]

Note:  

There are no electrons in a nucleus so, with the emission of a particle, one of the neutrons is converted to a proton and an electron.

[tex]_0n^1\ --------->\ _1P^1\ +\ _{-1}e^0[/tex]

(b) Positron Emission (Beta+ decay):

When an unstable nucleus decays by the emission of the positron, its charge number ‘Z’ decreases by 1 but,  its mass number ‘A’ remains unchanged. The transformation is represented by the equation:

[tex]_zX^A\ -------->\ _{Z-1}Y^A\ +\ _{+1}e^0[/tex]

Examples:

[tex]_{15}P^{30}\ -------->\ _{14}Si^{30}\ +\ _{+1}e^0[/tex]

Note:  

Inside the nucleus, only a proton can be transformed into a neutron with the emission of a positron (anti-particle of electron)

[tex]_1P^1\ -------->\ _0n^1\ +\ _{+1}e^0[/tex]

Answer:

Total energy, T= 4704 Joules

Explanation:

Given the following data;

Mass = 8kg

Initial height, h1 = 100m

Final height, h2 = 40 m

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the total energy, T;

T = mg(h1 - h2)

T = 8 * 9.8 * (100 - 40)

T = 78.4 * 60

Total energy, T= 4704 Joules

Farenheit is an interesting unit so the conversions get a bit weird.

To convert from Farenheit to Celcius you follow this formula:

T(°C) = (T(°F) - 32) × 5/9

It is also easier to convert to Celcius to Kelvin shown in this formula:

T(K) = T(°C) + 273.15

To Answer your question:

105° F = 40.5°C

-25° F = -31.7°C

105° F = 313.65K

-25° F = 241.45K

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = [tex]\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

[tex]\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }[/tex]

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

[tex]\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }[/tex]

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

[tex]\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }[/tex]

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

[tex]\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }[/tex]

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

[tex]\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }[/tex]

=1.67 x 20

t₁= 33.4   years

Answer:

the formula of mechanical advantage is

MA = load / effort

VR = effort distance / load distance

hope it is helpful to you

Answer:

A: An object has potential energy (stored energy) when it is not in motion. Once a force has been applied or it begins to move the potential energy changes to kinetic energy (energy of motion).

Explanation: