a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle down the slope is

Answer:

735 N

Explanation:

If a woman has mass 75kg and u know that mass is constant everywhere then just apply the formula W=mg...as gravity on earth is 9.8 m/s2 , so her weight will be 735 N...hope it helps...

If a woman has a mass of 75 kilograms then her weight on the earth would be 735.75 Newtons, because the weight of the woman is the multiplication of the mass of the woman and the acceleration due to the gravity of the earth.

It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another,

As given in the problem if a woman has a mass of 75 kilograms we have to find the weight of the woman,

The weight of the woman = mass of the woman × acceleration due to the gravity of the earth

                                           = 75 kilograms × 9.81

                                           =735.75 Newtons

Thus, the weight of a woman who has a mass of 75 kilograms would be 735.75 Newtons

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Answer:

A. The scale exerts a 17 N force up on the bag.

Explanation:

I Just took the test

Reaction force of the bag sitting on the scale a)The scale exerts a 17 N force up on the bag.

According to newton's third law of motion , every action have a equal and opposite reaction

When bag is being put on the scale to measure its weight , then bag must have exerted a force on the scale , in result of which an equal and opposite force must be exerted by the scale on the bag (according to newton's third law of motion) in order to keep an equilibrium state where both the forces are equal but opposite to each other .

correct option is a)The scale exerts a 17 N force up on the bag.

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Answer:

The current and the applied emf can be in phase if either of the two changes are made.

1) The inductance of the inductor is doubled, with everything else remaining constant.

2) The capacitance of the capacitor is doubled, with everything else remaining constant.

Explanation:

The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.

The phase difference between current and applied emf is given as

Φ = tan⁻¹ [(XL - Xc)/R]

XL = Impedance due to the inductor

Xc = Impedance due to the capacitor

R = Resistance of the resistor.

For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0

But only tan⁻¹ 0 = 0 rad

So, for the phase difference to be 0,

[(XL - Xc)/R] = 0

Meaning

XL = Xc

But for this question,

XL = 100 Ω, Xc = 200 Ω

For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.

The impedance are given as

XL = 2πfL

Xc = (1/2πfC)

f = Frequency

L = Inductance of the inductor

C = capacitance of the capacitor

The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.

And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.

So, these are either of the two ways to make the current and applied emf to be in phase.

Hope this Helps!!!

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle  and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

Explanation:

A metal such as copper is a conductor because it provides a pathway for electric charges to move easily. A material such as rubber is an insulator because it resists the flow of electric charges. A material that partially conducts electric current is a semiconductor. These materials include group 3 and group 5 elements.

Answer:

conductor

insulator

resists

semiconductor

group 3 and group 5

Explanation:

Answer:

The  z-component of the force is  [tex]\= F_z = 0.00141 \ N[/tex]    

Explanation:

From the question we are told that

          The charge on the particle is [tex]q = 7.76 *0^{-8} \ C[/tex]    

           The magnitude of the magnetic field is  [tex]B = 0.700\r i \ T[/tex]

            The  velocity of the particle toward the x-direction is  [tex]v_x = -1.68*10^{4}\r i \ m/s[/tex]

           The  velocity of the particle toward the y-direction is

[tex]v_y = -2.61*10^{4}\ \r j \ m/s[/tex]

           The  velocity of the particle toward the z-direction is

[tex]v_y = -5.85*10^{4}\ \r k \ m/s[/tex]

Generally the force on this particle is mathematically represented as

          [tex]\= F = q (\= v X \= B )[/tex]

So  we have    

          [tex]\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)[/tex]

         [tex]\= F = q (v_y B(-\r k) + v_z B\r j)[/tex]      

  substituting values

       [tex]\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)[/tex]    

      [tex]\= F= 0.00303\ \r j +0.00141\ \r k[/tex]                  

So the z-component of the force is  [tex]\= F_z = 0.00141 \ N[/tex]    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Answer:

Explanation:

initial vertical velocity = 17.5 m/s

using g=-9.81 m/s^2

apply kinematics equation

v1^2-v0^2=2gS

solve for S with v1=0, v0=+17.5

S = (v1^2-v0^2)/2g

=(0-17.5^2)/(2*(-9.81))

= 15.61 m

Answer:

the internal energy of the mixture at final state = 238kJ/kg

Explanation:

Given

V= 0.6m³

m=5kg

R=0.287kJ/kg.K

T=320 K

from ideal gas equation

PV = nRT

where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.

Recall, mole = mass/molar mass

attached is calculation of the question.

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

[tex]A_1v_1=nA_2v_2[/tex]               (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]

Then, the number of capillaries is 1.6*10^9

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

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Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.

Answer:

103 g

Explanation:

given that

Mass of the attached mass, m = 40 g

Initial distance of balance, x1 = 49.7 cm = 0.497 m

Final distance of balance, x2 = 39.2 cm = 0.392 m

Equilibrium point of attachment, x3 = 14 cm = 0.14 m

To get the mass of the meter stick, we use the relation

M = m.[(x2 - x3) / (x1 - x2)]

And on solving in full, we gave

M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]

M = 40(0.258 / 0.105)

M = 40 * 2.457

M = 102.8 g

Therefore the needed work good for mass of the stick j

Is 103 g

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

[tex]D = 0.5 c \rho A v^2[/tex]

Object 1 has twice the diameter of object 2.

If [tex]d_2 = d[/tex]

[tex]d_1 = 2d[/tex]

Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]

Area of object 1:

[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]

Since all other parameters are still the same except the drag coefficient:

For object 1:

[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]

For object 2:

[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]

Since the drag force for the two objects are the same:

[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

Answer:

The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

[tex]r=\frac{L}{2 \pi}\\[/tex]

  [tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]

area of the loop Is:

[tex]A_L= \pi r^2[/tex]

     [tex]=100.2001\times 3.14\\\\=314.628[/tex]

change in magnetic field is:

[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]

then the induced emf is:  [tex]e = A_L \times \frac{dB}{dt}[/tex]

                                              [tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]

resistivity of the copper wire is: [tex]\rho =[/tex]  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               [tex]= 0.5 \times 10^{-3} \ m[/tex]

hence the resistance of the wire is:

[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]

   [tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]

Power:

[tex]P=\frac{e^2}{R}[/tex]

[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]

The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]

Answer:

it is desired that the lenses stop this ray, its polarization must be vertical

Explanation:

To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.

The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.

Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is

                        n = tan teaP

Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

Answer: lower

There are a number of factors that can affect the coefficient of friction, including surface conditions.

Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively

Answer:

Seismic attenuation describes the energy loss experienced by seismic waves as they propagate. It is controlled by the temperature, composition, melt content, and volatile content of the rocks through which the waves travel.

Explanation:

Answer:

0.912

Explanation:

Given that

Height of bouncing of the ball, h = 1.71 m

Number of times the ball bounced, n = 4 times

Height from which the ball was dropped, H = 2.47

First, let's start by defining what coefficient of restitution means

Coefficient of Restitution, CoR is the "ratio of the final to initial relative velocity between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision."

It is mathematically represented as

CoR = (velocity after collision) / (velocity before collision)

1.71 = 2.47 * c^4, where c = CoR

1.71/2.47 = c^4

c^4 = 0.6923

c = 4th root of 0.6923

c = 0.912

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

b. only ray that reflects back in the same direction it came from

Explanation:

C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.

Correct answer is option B.

C-ray is the only ray that reflects back in the same direction it came from.

Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.

Answer:

4Hz (240 cycles/60 seconds = 4 cycles/second)

Explanation:

hope this helped!

Answer:

(a) Fn = 0.33 N

(b) Pn = 263.22 x 10³ Pa = 263.22 KPa

Explanation:

(a)

First, we need to calculate the total force exerted by all nails on the magician. This force must be equal to the weight of magician:

F = W

where,

F = Total Force exerted by all nails = ?

W = Weight of magician = mg = (64.3 kg)(9.8 m/s²) = 630.14 N

Therefore,

F = 630.14 N

Now, we calculate the force exerted by each nail:

Fn = F/n

where,

Fn = force exerted by each nail = ?

n = Total no. of nails = 1900

Therefore,

Fn = 630.14 N/1900

Fn = 0.33 N

(b)

The pressure exerted by each nail is given as:

Pn = Fn/An

where,

Pn = Pressure exerted by each nail = ?

An = Area of contact for each nail = 1.26 x 10⁻⁶ m²

Therefore,

Pn = 0.33 N/1.26 x 10⁻⁶ m²

Pn = 263.22 x 10³ Pa = 263.22 KPa

Answer:

It is the mean of 460 swimmers

Explanation:

In this question, we are concerned with knowing the mean of the population w

Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study

The population mean in this case is simply the mean of the swimming times of the 460 swimmers

There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study

So conclusively, the population mean w is simply the mean of the total 460 swimmers

Answer:

[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]

Explanation:

The gravitational force between two corpses is given by the following equation:

[tex]F = GMm/d^2[/tex]

Where F is the force, G is the gravitational constant

([tex]G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}[/tex]), M and m are the masses of the corpses and d is the distance between them.

So we have that:

[tex]F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2[/tex]

[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]

There are six statements on the list.

The first 2 are true, and the last 2 are true.

The 2 in the middle aren't true.  They are false.

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

t = 3.4 s

The box will come to rest in 3.4 s

Explanation:

For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma

Frictional Force = μR = μW = μmg

Therefore,

ma = μmg

a = μg

where,

a = acceleration of box = ?

μ = coefficient of sliding friction = 0.45

g = 9.8 m/s²

Therefore,

a = (0.45)(9.8 m/s²)

a = -4.41 m/s²  (negative sign due to deceleration)

Now, for the time to stop, we use first equation of motion:

Vf = Vi + at

where,

Vf = Final Speed = 0 m/s (since box stops at last)

Vi = Initial Speed = 15 m/s

t = time to stop = ?

Therefore,

0 m/s = 15 m/s + (-4.41 m/s²)t

(-15 m/s)/(-4.41 m/s²) = t

t = 3.4 s

The box will come to rest in 3.4 s

Answer:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.