Can someone please help, ty!!
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Answer:

The answer is 1.63 x 10 -7 N

Explanation:

An appropriate solution is 1.63 x 10 -7 N

Gravitational pressure method: F=G{frac{m_1m_2}{r^2}}

Newton's law of gravitation is Gm;m2

The gravitational regular G is 6.67 x 10-11 N·m2/kg2.

Definitions of gravitational pressure. The force of attraction among all masses inside the universe; especially the appeal of the earth's mass for bodies close to its floor.

Newton's regulation of typical gravitation states that two bodies in an area pull on every other with a force proportional to their hundreds and the space among them. For large objects orbiting each other—the moon and Earth, for instance—means they without a doubt exert important force on each other.

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Answer: C ON EDGE 2021

Explanation:

Answer:

el continuo movimiento

Explanation:

Answer:

Δϕ = -2.89 x 10^-4 Wb

Explanation:

Given that A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil

Let ΔB = Change in magnetic field

A = cross sectional area = 8.5 cm^2

t = time = 12ms

Flux change = Δϕ = (ΔB*A) / t

Δϕ =  [(0.02 - 0.06)x(8.5 x 10^-2)^2 ] / 0.012

Δϕ = (-0.04 x 7.225^-3) / 0.012

Δϕ = -2.89 x 10^-4 Wb

Therefore, the magnitude of the total change in the external magnetic flux enclosed by the coil is -2.89 x 10^-4 Wb

Analysing the Question:

Neglecting air resistance

We are given:

mass (m) = 15 kg

time taken by the rock to reach the water (t) = 1 second

height of the cliff (h) = h m

initial velocity of the rock (u) = 0 m/s  [ the rock was 'dropped' and not 'thrown' downwards]

acceleration of the rock due to gravity (a) = 9.8 m/s²

Finding the height to the cliff:

from the seconds equation of motion

s = ut + 1/2*at²

replacing the variables

h = (0)(1) + 1/2 * (9.8) *(1)²

h = 4.9 m

Therefore, the cliff is 4.9 m high

Answer:

your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force

Answer:

1.01 m/s

Explanation:

Given

initial speed, u = 1.8m/s

acceleration, a = -3.00m/s² (it's negative because the skater slowed down)

distance, s = 0.37m

Required

Determine the final velocity (v)

This will be solved using the following equation of motion

[tex]v\² = u\² + 2as[/tex]

Substitute values for u, a and s

[tex]v\² = 1.8\² + 2 * -3 * 0.37[/tex]

[tex]v\² = 3.24 - 2.22[/tex]

[tex]v\² = 1.02[/tex]

Take square roots

[tex]v = \sqrt{1.02[/tex]

[tex]v = 1.00995049384[/tex]

[tex]v = 1.01 m/s[/tex] (Approximated)

The final speed of the skater when she has slid 0.37 m is 1.0 m/s.

The given parameters:

The final speed of the skater when she has slid 0.37 m is calculated as follows;

[tex]v_f^2 = v_i^2 + 2ad\\\\v_f ^2 = 1.8^2 + 2(3)(-0.37)\\\\v_f^2 = 1.02\\\\v_f = \sqrt{1.02} \\\\v_f = 1.0 \ m/s[/tex]

Thus, the final speed of the skater when she has slid 0.37 m is 1.0 m/s.

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Answer:

33°

Explanation:

We are given;

Speed at which both snowballs are thrown; v = 33.2 m/s

Angle at which snowballs are thrown with respect to the horizontal; θ = 57°

Now,we want to find out the angle at which the second snowball should be thrown in order to arrive at the same point as the first.

To calculate this angle, we will use complementary angle concept.

Now, because the target is in the same place, there will be two launch angles that will make the snow ball to be placed on the target.

The is calculated from;θ1 = 45° - (57° - 45°) = 33°

Answer:

l think the answer is C. part

The degree to which the measurement approaches the quantity's true value

is what best describes the accuracy of a measurement.

When a quantity is measured, the value is usually unknown by the individual

taking the measurement . This should however be done with a very good

tool/equipment to eliminate inaccuracy.

The accuracy of a quantity is usually measured by how close the difference

is between the measurement taken and the actual value of the quantity.

The different  in value gotten should always be very small.

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Answer:

1) a = 6.14 km

2) b = 4.69 km

Explanation:

Let the first building be A, second building be B and third building be C.

Now, bearing of A = 4.76 km in a direction 37° north of east

Bearing of B = 69° west of north

Bearing of C = 28° east of south

Thus if this 3 points form a triangle, we will have the following angles;

Angle at point A = 28 + (90 - 37) = 81°

Angle at point B = 28 + (90 - 69) = 49°

Angle at point C = 180 - (81 + 49) = 50°

Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;

Using sine rule, we can find "a".

a/sin 81 = 4.76/sin 50

a = 6.14 km

B) distance between first and third building is AB in the triangle depicted by "b".

Similar to the first problem, we will use sine rule again.

b/sin49 = 4.76/sin 50

b = 4.69 km

Answer:

the farther you are from earth

Explanation:

It is this because when you get into space there is less gravity

Answer:

The time taken is  [tex]t =   6.102 \  hours [/tex]

Explanation:

From the question we are told that  

    The velocity of the train is [tex]v_t = 6.1 \ m/s[/tex]

    The distance covered by the train is  [tex]D =  134 \  km  =  134000 \  m[/tex]

Generally the time taken is mathematically represented as

     [tex]t =  \frac{  134000}{ 6.1 }[/tex]

=>    [tex]t =  \frac{  134000}{ 6.1 }[/tex]

=>    [tex]t = 21967.21 \ sec [/tex]

Converting to hours

       [tex]t = 21967.21  =  \frac{21967.21 }{3600} =   6.102 \  hours [/tex]

Answer:

Explanation:

[tex]a_{x}=0[/tex] [tex]v_{xo}= v_{o}cos(delta)t[/tex] [tex]a_{y}=-g[/tex] [tex]v_{yo}=sin[/tex]θ

X-direction                                     | Y-direction

[tex]x=x_o+v_{xo}t[/tex] ⇒ [tex]x=v_{xo}cos(delta)t[/tex] |

Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

Although the initial energy (potential energy is preserved), the energy of deformation as the ball strikes a surface creates energy dissipation in the form of frictional heat and audible sound energy.

Every time the ball bounces, its height will be less than its previous height.

Explanation:

Answer:

C is incorrect I got it wrong on my test don't go with C

Explanation:

Answer:

a) C = (5-4D)/3

Explanation:

Given the simultaneous equation

3C+4D = 5 .... 1

2C+5D = 2 .... 2

In order to isolate one of the variables, we will make one of the variables in any of the equation.

Using equation 1:

3C+4D = 5

Make C the subject of the formula:

Subtract 4D from both sides of the equation.

3C+4D-4D= 5-4D

3C = 5-4D

Divide both sides by 3:

3C/3 = (5-4D)/3

C = (5-4D)/3

Hence the expression that would result from the process is C = (5-4D)/3

Answer:

a

Explanation:

its correct

Answer:

The equilibrium temperature is 45°C

Explanation:

Given;

mass of block of ice, m₁ = 10 kg

mass of water, m₂ = 10 kg

initial temperature of the ice, t₁ = 0.0°C

initial temperature of water, t₂ = 90.0°C

let the equilibrium temperature = T

Apply the principles of mixture;

Heat lost by the water = heat gained by the ice

m₂c(90 - T) = m₁c(T - 0)

m₂(90 - T) = m₁(T - 0)

10(90 - T) = 10(T)

90 - T = T

90 = 2T

T = 90 / 2

T = 45°C

Therefore, the equilibrium temperature is 45°C

Answer:

Lower than it should be.

Explanation:

Hello.

In this case, since the density is computed by:

[tex]\rho =\frac{m}{V}[/tex]

And we obtain the volume of the solid by substracting the mass of the water and the solid minus the mass of water:

[tex]V_{solid}=V_{solid\ with \ water}-V_{water}[/tex]

If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.

Regards.

Answer:

7.0 m/s

Explanation: I just did it

Answer:

[tex]a=2.8\ m/s^2[/tex]

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2[/tex]

So, the acceleration of the object is [tex]2.8\ m/s^2[/tex].

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) High and low pressures in kilopascals:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

[tex]p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}[/tex]

[tex]p_{high} = 15.999\,kPa[/tex]

[tex]p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}[/tex]

[tex]p_{low} = 10.666\,kPa[/tex]

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures in pounds per square inch:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

[tex]p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}[/tex]

[tex]p_{high} = 2.320\,psi[/tex]

[tex]p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}[/tex]

[tex]p_{low} = 1.547\,psi[/tex]

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures in meter water column in meters water column:

We can calculate the equivalent water column of a mercury column by the following relation:

[tex]\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}[/tex]

[tex]h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}[/tex] (Eq. 1)

Where:

[tex]\rho_{w}[/tex], [tex]\rho_{Hg}[/tex] - Densities of water and mercury, measured in kilograms per cubic meter.

[tex]h_{w}[/tex], [tex]h_{Hg}[/tex] - Heights of water and mercury columns, measured in meters.

If we know that [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], [tex]h_{Hg, high} = 0.120\,m[/tex] and [tex]h_{Hg, low} = 0.080\,m[/tex], then we get that:

[tex]h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m[/tex]

[tex]h_{w, high} = 1.632\,m[/tex]

[tex]h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m[/tex]

[tex]h_{w, low} = 1.088\,m[/tex]

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Answer:

The area under the speed - time graph denotes the distance travelled by the object

In the given graph, we just have to think about the first 4 seconds, we also notice that the velocity at 4 seconds is 20 m/s

The distance travelled by the object in 4 seconds is the area of the triangle in the graph with a base of 4 units and height of 20 units (image included)

Distance Travelled = Area of triangle =  1/2 * base * height

Distance Travelled = 1/2 * 4 * 20

Distance travelled = 4 * 10

Distance travelled = 40 m

Answer:

D. Acceleration remains constant and velocity is negative and decreasing.

Explanation:

Khan Academy

Explanation:

1) 1 : to Prevent a burning we use tongs

2:it can burn our skin hoa it's help

2) 1: because pulley helps in water draw from a well

3) 1: second class lever has mechanical advantage more than one as load is in between fulcrum an effort making the effort arm longer than the load arm

Answer:

the mass and wieght means same

Answer:

Find the answer in the explanation.

Explanation:

Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.

A.) Since the radius of the earth is equal to 6400 km. At it to 400km

The radius of the orbit = 6400 + 400

The radius = 6800 km.

Where orbital speed = sqrt ( g × r )

Orbital speed = sqrt ( 6800 × 8.69 )

Orbital speed = 243.09 km/h

B.) Orbital period acceleration to Kepler third law of planetary motion state that:

The square of the period is directly proportional to the cube of the radius.

T^2 = (4π^2 /GM) r^3

T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3

T^2 = (4π^2/ 4.002^36) × 6800^3

T^2 = 9.865×10^-36 × 6800^3

T^2 = 3.10 × 10^-15

T = 5.57 × 10^-8 hours.

Convert the hours to minutes by multiplying it by 60

T = 3.4 × 10^-6 minutes.

Since the radius of the earth is equal to 6400 km. At it to 400km

The radius of the orbit = 6400 + 400

The radius = 6800 km.

Where:

The speed of the ISS is 243.09Km/h.

Part B:

The orbital period (T) of the ISS in minutes is :

Convert the hours to minutes by multiplying it by 60

T = 3.4 × 10^-6 minutes.

The orbital period (T) of the ISS in minutes is 3.4 × 10^-6 minutes.

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Answer:

h = 219.96 m

Explanation:

Speed of the stone with which it was thrown horizontally, v = 10 m/s

We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.

It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0

Using second equation of motion,

[tex]d=u't+\dfrac{1}{2}at^2[/tex]

Put u'=0 and a=g

So,

[tex]d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m[/tex]

So, the height of the cliff is 219.96 m.