Building helps children develope their _____ skills

A. number sense and operations

B. operations and geometry

C. measurement and counting

D. measurement and geometry

Answer:

It means at a diverging boundary when magma spreads to both sides it is almost identical in both the side.

Explanation:

At a diverging boundary when magma spreads to both sides it is almost identical in both the side.

If you take a picture of one side of the boundry and the spreading of sea floor and place it before a mirror you can see the image is identical to the picture of other side.

Therefore, the meaning of saying, each side of the sea floor away from the mid ocean ridge is a mirror image of the other side is explained above.

The statement that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side refers to the symmetric nature of seafloor spreading. Molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust, creating symmetric magnetic striping patterns. This is akin to the bilaterally symmetric structure observed in certain organisms.

When we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side, we are referencing the symmetric nature of seafloor spreading. Just as a mirror image reflects an object exactly, so does the seafloor on one side of the mid-ocean ridge reflect the seafloor on the opposite side. This is because molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust. This new crust then moves away from the ridge due to tectonic forces, and the process repeats, creating a pattern of symmetrical magnetic striping on the seafloor.

This 'mirroring' effect is similar to the bilaterally symmetric structure seen in certain organisms, where a plane cut from the front to back of the organism produces distinct left and right sides that are mirror images of each other. You can see this symmetry in the images of the Moon provided – two different sides, yet mirroring similar physical traits.

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Answer:

39.8 m

Explanation:

Let h be the required height and t be the time to reach the egg which was thrown later on. The egg thrown earlier took ( t +2.2) s , with initial velocity zero to cover distance of h.,

So

h = 1/2 g( t +2.2)²

For the egg thrown with velocity of 58 m/s

h = 58 t + 1/2 g t²

Equating these two equations

1/2 g( t +2.2)²  =  58 t + 1/2 g t²

1/2 g ( t² + 4.4 t + 4.84 ) = 58 t + 1/2 g t²

58 t = 21.56 t + 23.716

t = .65 s

h = 1/2 x 9.8 x (2.2+.65)²

h = 39.8 m

Answer:

a) 0.063 m/s

b)  243.83 m/s

Explanation:

given,

distance = 7900 km

time = 4 years

a) average velocity for the trip = [tex]\dfrac{7900}{4\times 365}[/tex]

                                                   = 5.41 km/day

                                                   = [tex]\dfrac{5.41\times 1000}{24\times 60\times 60}[/tex]

        average velocity for the trip = 0.063 m/s

b) average velocity  = [tex]\dfrac{7900\times 1000}{9\times 60\times 60}[/tex]

average velocity for the trip = 243.83 m/s

Answer:

p = 727.273 kg/m3,  y = 7134.545 N/m^3, SG = 0.7273

Explanation:

Density is simply the amount of mass of a substance per unit of volume. It can be found by dividing the mass in kg by the volume im m^3:

[tex]p = \frac{m}{v}  = \frac{40kg}{55lt*\frac{1 m^3}{1000 lt}} = 727.273 kg/m^3[/tex]

Specific weight is the weight of the substance per unit of volume. The weight is the mass of the material times the gravity, and it represents the force that the earth exerts on an object. Another way of calculate this value, its multiplying the density of the fuel times the gravity. Then:

[tex]y =  p*g = 727.273 kg/m^3 * 9.81 m/s^2 = 7134.545 N/m^3[/tex]

Specific Gravity is the ratio of the density of the substance to the density of a reference substance. For liquids, the reference substance is water at 4°C, which has a density of about 1000 kg/m^3.

[tex]SG =\frac{ p_{fuel}}{p_{water}}  = \frac{727.273 kg/m^3}{1000 kg/m^3} = 0.7273[/tex]

Answer:d

Explanation:

Given

Magnitude of Vector A is 3 m

and Magnitude of vector B is 4 m

So the maximum value of resultant can be 7 m when both are at an angle of [tex]0^{\circ}[/tex]

and its minimum value can be 1 m when both are at angle of [tex]180 ^{\circ}[/tex]

So the resultant must lie between 1 m to 7 m  

The length of the sum of the vectors A and B with no direction given must be some value from 1 m to 7 m. Thus, the correct option is D.

Vector addition is the property or operation of addition of two or more vectors together into a vector sum. The sign used with the vectors depends upon the direction in which they move. If the vectors are moving in the same direction then they will be added and if they are moving in opposite directions then they are substracted.

Given, Magnitude of Vector A is 3 m and Magnitude of vector B is 4 m.

So, in this case the maximum value of resultant vector can be 7 m when both are moving in the same direction with an angle of 0° and it has minimum value which can be 1 m when both are at angle of 180° in opposite direction.

So, the resultant vector must lie between 1 m to 7 m.  

Therefore, the correct option is D.

Learn more about Vector sum here:

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Answer:

a) [tex]y_{max}=7.10m[/tex]

b) [tex]t=2.40 s[/tex]

Explanation:

From the exercise, we know the initial velocity, gravitational acceleration and initial position of the dolphin.

[tex]v_{oy}=11.8m/s[/tex]

[tex]y_{o}=0m\\ g=9.8m/s^{2}[/tex]

a) To find maximum height, we know that at that point the dolphin's velocity is 0 and it becomes coming down later.

Knowing that, we need to know how much time does it take the dolphin to reach maximum height.

[tex]v_{y}=v_{oy}+gt[/tex]

[tex]0=11.8m/s-(9.8m/s^{2} )t[/tex]

Solving for t

[tex]t=1.20 s[/tex]

So, the dolphin reach maximum point at 1.20 seconds

Now, using the equation of position we can calculate maximum height.

[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]

[tex]y=0+11.8m/s(1.20s)-\frac{1}{2}(9.8m/s^{2} )(1.20s)=7.10m[/tex]

b) To find how long is the dolphin in the air we need to analyze it's hole motion

At the end of the jump the dolphin return to the water at y=0. So, from the equation of position we have that

[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]

[tex]0=0+11.8t-\frac{1}{2}(9.8)t^{2}[/tex]

What we have here, is a quadratic equation that could be solve using:

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-\frac{1}{2} (9.8)[/tex]

[tex]b=11.8\\c=0[/tex]

[tex]t=0s[/tex] or [tex]t=2.40 s[/tex]

Since the answer can not be 0, the dolphin is 2.40 seconds in the air.

Answer:14.72 m/s

Explanation:

Given

Initial velocity (u)=16.6 m/s

[tex]\theta =40.9^{\circ}[/tex]

Horizontal velocity component ([tex]u_x[/tex])=16.6cos40.9=12.54 m/s

As the ball comes down so its vertical displacement is zero except 3 m elevation

Thus [tex]v_y=\sqrt{\left ( 16.6sin40.9\right )^2+2\left ( -9.81\right )\left ( 3\right )}[/tex]

[tex]v_y=\sqrt{10.868^2-58.86}[/tex]

[tex]v_y=\sqrt{59.253}[/tex]

[tex]v_y=7.69 m/s[/tex]

there will be no change is horizontal velocity as there is no acceleration

Therefore Final Velocity

[tex]v=\sqrt{u_x^2+v_y^2}[/tex]

[tex]v=\sqrt{12.54^2+7.69^2}[/tex]

v=14.72 m/s

Answer:

D. Equalize voltage

Explanation:

Of the following lead-acid  battery the battery with voltage value is Equalize voltage. EQUALIZING lead acid battery is process of de-sulphating the battery by carrying out a controlled overcharge. When battery plates acquire sulphate coating over time, their efficiency reduces, by overcharging this sulpahte coating is blown-off and the battery is rejuvenated.

The question is asking to calculate the required minimum deceleration of a speeding car to fit the speed limit when it crosses the end of a measured stretch of the road. The deceleration required can be calculated using the motion equations of physics by first calculating the time it takes to travel the distance between the two strips and then using this time to calculate the required deceleration.

The subject area of this question is kinematics, which is a branch of physics that deals with the concepts of distance, displacement, speed, velocity, and acceleration. The problem is asking for the minimum deceleration the car needs to have in order to not exceed the speed limit when it reaches the second strip. To solve this, you would use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is acceleration (in this case deceleration, so it will be negative), and t is time. But first, we need to figure out the time. The time it takes to travel between the two strips can be calculated by t = d/v. The distance d is given as 110m, and the average speed v we want is 21 m/s (the speed limit).

Once we have the time, we can substitute the values into the first equation and solve for a. The values are v = 21 m/s (the speed we want to have in the end), v0 = 33 m/s (the initial speed), and we've calculated t from the previous step. For deceleration, as the speed is decreasing, a will be a negative number.

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Answer:

They have a difference in energy of 35 eV.

Explanation:

The energy at rest of a particle is given by:

[tex]E_{R} = m_{0}c^2[/tex]   (1)

Where [tex]m_{0}[/tex] is the mass of the particle at rest and c is the speed of light.

Beta particles are high energy and high velocity electrons or positrons ejected from the nucleus of an atom as a consequence of a radioactive decay. Either if the beta particle is an electron¹ or a positron² it will have the same mass.

Hence, the mass of the beta particle at rest in equation (1) will be equal to the mass of an electron:

[tex]m_{e} = 9.1095x10^{-31} Kg[/tex]

Replacing the values of [tex]m_{e}[/tex] and c in equation (1) it is gotten:

[tex]E_{R} = (9.1095x10^{-31} Kg)(3.00x10^{8} m/s)^{2}[/tex]

[tex]E_{R} = 8.19x10^{-14} Kg.m^{2}/s^{2}[/tex]

But [tex]1 J = Kg.m^{2}/s^{2}[/tex], therefore:

[tex]E_{R} = 8.19x10^{-14} J[/tex]

It is better to express the rest energy in electronvolts (eV):

[tex]1eV = 1.60x10^{-19} J[/tex]

[tex]8.19x10^{-14} J . \frac{1 eV}{1.60x10^{-19} J}[/tex] ⇒ [tex]511.875 eV[/tex]

[tex]E_{R} = 511.875 eV[/tex]

So the energy of the beta particle at rest is 511.875 eV.

Case for the one traveling at 0.35c:

Since it is traveling at 35% of the speed of light it is necessary to express equation (1) in a relativistic way, that can be done adding the Lorentz factor to it:

[tex]E = \frac{m_{0}c^{2}}{sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]   (2)

Where v is the velocity of the particle (for this case 0.35c).

[tex]E = \frac{511.875 eV}{sqrt{1-\frac{(0.35c)^{2}}{c^{2}}}}[/tex]

[tex]E = \frac{511.875 eV}{sqrt{1-\frac{0.1225c^{2}}{c^{2}}}}[/tex]

[tex]E = \frac{511.875 eV}{sqrt{1-0.1225}}[/tex]

[tex]E = \over{511.875 eV}{sqrt{0.8775}}[/tex]

[tex]E = \over{511.875 eV}{0.936}[/tex]

[tex]E = 546.875 eV[/tex]

The difference in energy between the two particles can be determined using the relativistic form of the kinetic energy:

[tex]K = E – E_{R}[/tex]  (3)

Where E is the energy of the particle traveling at 0.35c and [tex]E_{R}[/tex] is the energy of the beta particle at rest.

[tex]K = 546.875 eV – 511.875 eV[/tex]

[tex]K = 35 eV[/tex]

They have a difference in energy of 35 eV.

Key terms:

¹Electron: Fundamental particle of negative electric charge.

²Positron: Is an electron with positive electric charge (similar to an electron in all its properties except in electric charge and magnetic moment).

Answer:

[tex]W=2eV[/tex]

Explanation:

Energy from a light source (photons):

E=h*c/λ

Photoelectric effect and Work function (W):

E=W+Ek

Ek: maximum kinetic energy from photoelectrons

then:

h*c/λ=W+Ek

Case 1:

[tex]h*c/lambda_{1}=W+Ek_{1}[/tex]  (1)

Case 2:

[tex]h*c/lambda_{2}=W+Ek_{2}[/tex]

but  [tex]lambda_{2}=lambda_{1}/2[/tex]

[tex]2h*c/lambda_{1}=W+Ek_{2}[/tex]  (2)

If we divide (2) by (1):

[tex]2=\frac{W+Ek_{2}}{W+Ek_{1}}[/tex]

[tex]W=Ek_{2}-2Ek{1}=2eV[/tex]

Answer:

Force, [tex]F=8.71\times 10^{-4}\ N[/tex]

Explanation:

Given that,

Charges on pith balls, [tex]q_1=q_2=-28\ nC=-28\times 10^{-9}\ C[/tex]

Distance between balls, d = 9 cm = 0.09 m

Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :

[tex]F=k\dfrac{q_1^2}{d^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}[/tex]

F = 0.000871 N

or

[tex]F=8.71\times 10^{-4}\ N[/tex]

So, the repulsive force between the pith balls is [tex]8.71\times 10^{-4}\ N[/tex]. Hence, this is the required solution.

Answer:

The speed of the ball when it hits the ground is 83.4 m/s

Explanation:

Please see the attached figure for a description of the problem.

The vector velocity at time "t" can be written as follows:

[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]

where:

v0 : module of the initial velocity vector.

α: launching angle.

g: acceleration due to gravity.

t: time

To calculate the impact speed, we can use this equation once we know the time at which the ball hits the ground. For this, we can use the equation for position:

[tex]r = ( x0 + v0*t*cos\alpha ; y0 + v0*t*sin\alpha + 1/2 g*t^{2})[/tex]

where:

r= vector position

x0 = horizontal initial position

y0 = vertical initial position

The problem gives us information about the vertical displacement. If we take the base of the wall as the center of the reference system, at the time at which the ball hits the ground, the module of the vertical component of the vector position, r, is 0 m (see figure). The initial vertical position is  22 m.

if ry is the vertical component of the vector r at final time:

[tex]ry = (0; y0 + v0*t*sin\alpha +1/2 g*t^{2})[/tex]

then, the module of ry is (see figure):

module ry =[tex]0 = 22m + v0*t*sin\alpha + 1/2*g*t^{2}[/tex]

Let´s replace with the given data:

[tex]0 = 22m + 39.3 m/s*t -4.9 m/s^{2}*t^{2}[/tex]

Solving the quadratic equation:

t = -0.5 and t = 8.5 s

At 8.5 s after firing, the ball hits the ground.

Now, we can find the module of the velocity vector when the ball hits the ground:

[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]

at time t = 8.5s

[tex]v = (81.0m/s * cos\alpha ; 81.0m/s * sin\alpha - 9.8 m/s^{2} *8.5s)[/tex]

v = (70.8 m/s; -44.0 m/s)

module of v = [tex]\sqrt{(70.8m/s)^{2} + (-44.0m/s)^{2}}[/tex] = 83.4 m/s

Answer:

V=120m/s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t                                     (1)

{Vf^{2}-Vo^2}/{2.a} =X                (2)

X=Xo+ VoT+0.5at^{2}                   (3)

X=(Vf+Vo)T/2                                 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem you must divide the problem into two parts 1 and 2, when the rocket accelerates (1), and when the rocket decelerates (2).

Then you raise equations 3 and 1 in both parts.

finally you use algebraic methods to find the value of speed

I attach the complete procedure

Answer:169.1 N

Explanation:

Given

bird weigh 38 pounds

and we know

1 Pound is equal to 0.453592 kg

Thus 38 pounds is equal to 17.236 kg

Thus the weight of bird is [tex]17.236\times 9.8=169.09 N \approx 169.1 N[/tex]

Answer:

The weight in newtons is 169.1.

Explanation:

This question can be solved as a simple rule of three problem.

We have that each pound has 4.45 newtons. So, how many newtons are there in 38 pounds?

So

1 pound - 4.45N

38 pounds - xN

[tex]x = 38*4.45[/tex]

[tex]x = 169.1N[/tex]

The weight in newtons is 169.1.

Answer:

if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.

Y (position)

I

sensor-------------------------------------------------------------------

I                                                    o

I                                     o

I                           o

I                   o

I            o

I       o

I   o

I------------------------------------------------------------------------------------- X (time)

in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.

Y (position)

I

sensor ----------------------

I       o

I                 o

I                          o

I                                 o

I                                       o

I                                            o

I                                              o

I------------------------------------------------------------------------------------- X (time)

i hope you can understand it, kinda hard to do graphs here.

Answer:

news will reach to the listener who are 42 km apart

Explanation:

given,

distance of the radios from the station = 42 Km

speed of the sound in the air = 343 m/s

distance of the people = 9.26 m

[tex]time =\dfrac{distance}{speed}[/tex]

time taken by the signal to reach to the radio

speed of the electromagnetic wave to reach to the people

speed of electromagnetic wave = 3 × 10⁸ m/s

[tex]t =\dfrac{42000}{3 \times 10^8}[/tex]

t = 1.4 μs

time taken to reach to the people

[tex]time =\dfrac{distance}{speed}[/tex]

[tex]t =\dfrac{9.26}{343}[/tex]

t = 27 ms

time taken by the station to radio is less.

hence, news will reach to the listener who are 42 km apart

Answer:

1) 1001

=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 8 + 1 = (9)₁₀

2) 10101

=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 16 + 4 + 1 = (21)₁₀

3) 1010001

=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 64 + 16 + 1 = (81)₁₀

4) 1010.1010

= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴

= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀

Answer:

1) 1001

=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 8 + 1 = (9)₁₀

2) 10101

=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 16 + 4 + 1 = (21)₁₀

3) 1010001

=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰

= 64 + 16 + 1 = (81)₁₀

4) 1010.1010

= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴

= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀

Explanation:

Answer:

[tex]4.50*10^8\frac{N}{C}[/tex]

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

[tex]E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}[/tex]

The magnitude of the electric field created by a larger charge Q at the location of a test charge is calculated by the formula E = k*Q/r² where k is Coulomb's constant, Q is the charge creating the field, and r is the distance between the charges. Substituting the given values into the formula, we find the electric field magnitude to be 4.495 x 10^8 N/C.

The magnitude of the electric field created by a point charge Q can be calculated using Coulomb's law as follows:

The equation for calculating the electric field E in relation to the force F imparted on a small test charge q is defined as E = F/q.

However, the Coulomb’s law gives the force F between two charges as F = k*Q*q/r², where, k is the Coulomb's constant (8.99 × 10^9 N.m²/C²), Q is the charge creating the field, q is the test charge, and r is the distance between them.

In the case where we want to find the electric field created by a larger charge Q at the location of a small charge q, we consider the force on the small charge exerted by Q, and therefore, we rewrite the equation as E = k*Q/r².

Therefore for your question, the magnitude of the electric field (E) at the location of the test charge is calculated as E = (8.99 x 10^9 N.m²/C² * 5.0 C) / (10m)² = 4.495 x 10^8 N/C.

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Answer: A) we can write a wave packet like Y = ∑aₙ*cos(w*n*t)

So you have lots of different wavelengths here, and you think that a wave packet has a unified frequency, but not wavelength.

B) mathematically speaking you will write a wave packet like i did up there, has a sum of many waves, where for each one you can have an intensity aₙ and a different frequency, the only thing you must see is that all the waves you are suming are moving in the same direction.

Answer:0.558 m

Explanation:

Given

weight of Plank=228 N

weight of man=449 N

Length of plank=4.4 m

let [tex]R_a[/tex] and [tex]R_b[/tex] be the reactions at A & B

and Reaction at A becomes zero when plank is about to rotate at B

Taking moment about B at that instant so that plank is just about to rotate.

[tex]228\times 1.1=449\times x[/tex]

Where x is the maximum distance that man can walk

x=0.558 m

Answer:

J = 14.4 kg*m^2

Explanation:

Assuming that the wheel is not moving anywhere, and the kinetic energy is only due to rotation:

Ek = 1/2 * J * w^2

J = 2 * Ek / (w^2)

We need the angular speed in rad / s

566 rev/min * (1 min/ 60 s) * (2π rad / rev) = 58.22 rad/s

Then:

J = 2 * 24400 / (58.22^2) = 14.4 kg*m^2

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

t = 9.96 s

The sled required almost 10 s to travel down the slope.

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s[/tex]

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m[/tex]

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

Final answer:

The average speed for the trip to the hospital, which covers a distance of 6.0 miles in 15 minutes, is 24 miles per hour (mph).

Explanation:

To calculate the average speed of the trip to the hospital, you take the total distance traveled and divide it by the total time taken. The student has specified that it takes 15 minutes to drive 6.0 miles in total. Therefore, the average speed can be calculated as follows:

Speed = Total Distance / Total Time
Speed = 6.0 miles / 15 minutes
Speed = 0.4 miles per minute

To convert this to miles per hour (mph), multiply by 60 (since there are 60 minutes in an hour):

Speed = 0.4 miles/minute × 60 minutes/hour
Speed = 24 mph.

Therefore, the average speed for the entire trip is 24 mph.

Answer:

As from the given question,

The free body diagram of a ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground is show below.

Friction force can be calculated by formula:

Friction force = μ N

As we know that

At equilibrium,

The sum of the total horizontal force is equal to '0'.

The sum of the total vertical force is equal to '0'.

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

[tex]d\times sin\theta=m\times \lambda[/tex]

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, [tex]\theta[/tex]  = 15.8°

First bright fringe means , m = 1

So,

[tex]d\times sin\ 15.8^0=1\times \597\ nm[/tex]

[tex]d\times 0.2723=1\times \597\ nm[/tex]

[tex]d=2192.43481\ nm[/tex]

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

Distance between slits ≅ 2.2 µm

Answer:

The separation between the slit is [tex]2.19\mu m[/tex]

Solution:

As per the question:

Wavelength of light, [tex]\lambda = 597 nm = 597\times 10^{-9} m[/tex]

[tex]\theta = 15.8^{\circ}[/tex]

Now, by Young's double slit experiment:

[tex]xsin\theta = n\lambda[/tex]

here,

n = 1

x = slit width

Therefore,

[tex]x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m[/tex]

[tex]x = 2.19\mu m[/tex]

Answer:

Force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times

Explanation:

According to law of gravitation we know that force between two object is given by

[tex]F=\frac{GM_1M_2}{R^2}[/tex], here G is universal gravitational constant [tex]M_1[/tex] is mass of one object and [tex]M_2[/tex] is mass of other object , and R is the distance between them

From the relation we can see that force is inversely proportional to square of distance between them

So force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times

Final answer:

The strength of the gravitational force between the Earth and the Sun would decrease by a factor of approximately 6.30 if the distance between them were increased by a factor of 2.51, according to the inverse square law of universal gravitation.

Explanation:

If the distance between the Earth and the Sun were increased by a factor of 2.51, the strength of the force between them would change according to Newton's law of universal gravitation. This law states that the force is inversely proportional to the square of the distance between two masses. Therefore, if the distance increases by a factor of 2.51, the force of attraction would decrease by a factor of (2.51)2.

To calculate the specific factor by which the force decreases, we square 2.51 which gives us 6.3001. The gravitational force would thus be reduced by this factor. Since the law involves an inverse square relationship, as the distance increases by a factor of 2.51, the strength of the gravitational force decreases by a factor of approximately 6.30.

Answer:

Explanation:

length L = 5±.1

percentage error

= 0.1/5 x 100 = 2%

breadth b = 3.8 ±.1

percentage error

= .1 / 3.8 x 100

= 2.6 %

Maximum reading of length = 5.1 m

maximum reading of breadth = 3.9m

maximum area possible = 5.1 x 3.9

= 19.9 m².

maximum possible error in the measurement of area in percentage terms

= 2+2.6 = 4.6 %

error in maximum area

19.9 x 4.6 /100

= .9 m²

Minimum possible measurement of area

4.9 x 3.7 = 18.1 m²

possible error

18.1 x 4.6 / 100

= .8 m²

Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100)  x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %