Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N, and the angle θ between the strings coming from the two scales is 121 degrees.

Answer:

7.0 m/s

Explanation: I just did it

Answer:

B

Explanation:

B is Newton's first law of motion, I got this in class and that was the right answer.

Good Luck!

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

Answer:

1. Scalar

2.Vector

3. Scalar

4. Vector

5.Scalar

6.Scalar

7.Vector

8.Vector

9.Scalar

10.Scalar

11.Scalar

12. Vector

13.Scalar

Explanation:

Scalar refers to magnitude, and Vectors include magnitude with directions.

Answer:

el continuo movimiento

Explanation:

Answer:

C

Explanation:

Chemical energy is in the food you eat and It fuels your body

Answer: B

Explanation:

When you chew it, you're changing the physical state it's in.

Answer:

the mass and wieght means same

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:

Okay, first off, the formula for Kinetic Energy is:

KE = 1/2(m)(v)^2

m = mass

v = velcoity (m/s)

Using this formula, we can then calculate the kinetic energy in each scenario:

1) KE = 1/2(100)(5)^2 = 1,250 J

2) KE = 1/2(1000)(5)^2 = 12,500 J

3) KE = 1/2(10)(5)^2 = 125 J

4) KE = 1/2(100)(5)^2 = 1,250 J

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:

S=d/t.

distance is= speed multiplied by the time

D= 115 multiplied by 20

the answer is 2300km

Driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

To determine the distance the driver travels 20 seconds after receiving the text message, we need to calculate the distance covered during that time interval.

First, we convert the driver's speed from km/h to m/s for consistent units. The conversion factor is 1 km/h = 0.27778 m/s.

Driver's speed = 115 km/h × 0.27778 m/s = 31.944 m/s

Next, we use the formula for distance traveled:

Distance = Speed × Time

Distance = 31.944 m/s × 20 s = 638.88 meters

Therefore, the driver will be 638.88 meters away from the point where he received the text message after 20 seconds.

To convert this distance to kilometers, we divide by 1000:

Distance = 638.88 m ÷ 1000 = 0.63888 km

So, the driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

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Answer:

graph

noun

a diagram representing a system of connections or interrelations among two or more things by a number of distinctive dots, lines, bars, etc.

Mathematics.

a series of points, discrete or continuous, as in forming a curve or surface, each of which represents a value of a given function.

Also called linear graph . a network of lines connecting points.

a written symbol for an idea, a sound, or a linguistic expression.

verb (used with object)

Mathematics. to draw (a curve) as representing a given function.

to represent by means of a graph.

Explanation:

that is what i found

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:

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:

a

Explanation:

its correct

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.

Answer:

bc I'm ugly

Explanation:

i just answer a random thing bc I know is not from school

Answer:

you shouldn't, if u were to choose me then you would be making a huge mistake since choosing me would go against why you shouldn't choose me.

Answer:

frequency.

......... i need more characters so

Answer:

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

Answer:

a)     E = k Q / r² ,      k = 1 / 4πε₀,  b)  E=0, C) E=0,   d) Q=0

Explanation:

To find the electric field in this exercise let's use Gauss's law

         Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

Let's use a sphere as a Gaussian surface, in this case the field lines and the radii of the spheres are parallel so the scalar product is reduced to the algebraic product, let's carry out the integral

          E A = q_{int} /ε₀

The area of ​​a sphere is

         A = 4π r²

we substitute

          E (4π r²) =q_{int} /ε₀

a) The field is requested outside the spherical shell

        d> R₂

in this case the charge inside the Gaussian surface is

          q_{int} = Q

           E = Q / (4π r²) ε₀

           E = k Q / r²

           k = 1 / 4πε₀

b) In the second case, the field inside the spherical casing is requested.

As the surface is metallic, the charge is located on the surface of it, the electrons repel each other. So the charge inside is zero

           q_{int} = 0

            E = 0

c) The field in the interior region of the shell, all the charge is external, therefore the internal charge is zero

           q_{int} = 0

           E = 0

d) as the shell is conductive, the electrons can move freely, which is why it moves as far away as possible between them, this means that everything is located as close as possible to the outer surface of the conductor. Consequently, THERE ARE NO ELECTRONS ON THE INTERNAL SURFACE

              Q_internal =0

Answer:

A

 [tex]k = 0.11 \ m [/tex]  beyond the goal  post

B

Given that the goal  was score against the Bruins it then means that the Flyer won the game with 2 points

Explanation:

From the question we are told that

  The amount of time left in the game is  [tex]t =  20 \ s[/tex]

   The  amount of time the goal was defended  is  [tex]t_1  =  9 \  s[/tex]

   The  time remaining when the shot was made is  [tex]t_2 =  1.25 \  s[/tex]

     The  distance of the pluck from the goal is  [tex]d =  37 \  m[/tex]

     The  initial  velocity of the pluck is [tex]u =  30 \  m/s[/tex]

    The acceleration of the pluck is  [tex]a =  -0.50 \  m/s^2[/tex]

Generally the distance travel at the remaining time is mathematically represented as

      [tex]D =  u *  t _2  +  \frac{1}{2}* a*  t_2[/tex]

      [tex]D =  30 * 1.25  - \frac{1}{2}* 0.50* 1.25^2[/tex]

       [tex]D =  37.11 \ m [/tex]

So the position of the pluck when the game clock reaches zero is

        [tex]k = D - d[/tex]

=>    [tex]k = 37.11 - 37[/tex]

=>        [tex]k = 0.11 \ m [/tex]  beyond the goal  post

Given that the goal  was score against the Bruins it then means that the Flyer won the game with 2 points

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

[tex]v^2-c^2=\dfrac{w^2}{t^2}[/tex]

Put the value in the equation

[tex]6.00^2-c^2=\dfrac{w^2}{(20.1)^2}[/tex]

[tex] 36-c^2=\dfrac{w^2}{404.01}[/tex]....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

[tex]v^2-c^2=\dfrac{w^2}{t^2}[/tex]

Put the value in the equation

[tex]9.00^2-c^2=\dfrac{w^2}{(11.2)^2}[/tex]

[tex] 81-c^2=\dfrac{w^2}{125.44}[/tex]....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

[tex]45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}[/tex]

[tex]45=w^2(0.00549)[/tex]

[tex]w^2=\dfrac{45}{0.00549}[/tex]

[tex]w=\sqrt{\dfrac{45}{0.00549}}[/tex]

[tex]w=90.5[/tex]

We need to calculate the current speed

Using equation (I)

[tex]36-c^2=\dfrac{(90.5)^2}{(20.1)^2}[/tex]

[tex]36-c^2=20.27[/tex]

[tex]c^2=20.27-36[/tex]

[tex]c=\sqrt{15.73}[/tex]

[tex]c=3.96\ m/s[/tex]

(b). We need to calculate the shortest time

Using formula of time

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{90.5}{6}[/tex]

[tex]t=15.0\ sec[/tex]

We need to calculate the distance

Using formula of distance

[tex]d=vt[/tex]

[tex]d=3.96\times15.0[/tex]

[tex]d=59.4\ m[/tex]

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Answer:

a)

b is at higher potential

b)

E = 800 V/m

c)

W = -48.0 μ J

Explanation:

Given:

electric field has magnitude E and is directed in the negative x direction

potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) = 240 V

a)

Direction of the electric field is in the negative x direction so it is away from the positive charge and the more distance away from direction, the lesser the potential gets. So b is at the higher potential.

b)

When the magnitude of electric field is constant then the potential difference between two points in the field is given by:

Vab = Ed

where

Vab is the potential difference between two points a and b in the field

E is electric field magnitude

d is the distance between two points

Compute distance between two points a and b

d = 0.90 - 0.60

d = 0.30

d = 0.3 m

Since potential difference between two points in the field is given which is 240 V

Compute E:

The above formula becomes:

Vab = Ed

E = Vab/d

  = 240/0.3

E = 800 V/m

c)

The negative charge moved from the higher potential to low so work done on the point charge by electric field is negative.

W = Fd

Electric field = E = F/q

where

E is electric field magnitude

F is electric force on q

q is point charge magnitude

The charge q is either positive or negative and when charge is positive the directions of E and F are same otherwise opposite in case of negative charge.

So to calculate the work done on the point charge by the electric field:

E = F/q

F = Eq

so

W = Fd = Eqd

Now putting the values:

E = 800 V/m

q = - 0.200 μC

d = 0.3 m

W = Eqd

W = 800 * - 0.200 * 0.3

W = 800 * - 0.2 * 10⁻⁶ * 0.3

W = - 48 * 10⁻⁶

W = - 4.8 * 10⁻⁵ J

W = -48.0 μJ

a. The point which is at a higher potential is point a.

b. The magnitude (E) of the uniform electric field is equal to 800 V/m.

c. The work done on the point charge by the electric field is equal to -48 microjoules.

Given the following data:

a. Since the uniform electric field with a magnitude (E) is directed in the negative x direction, the point at a higher potential is point a because the electric field is not close to the positive charge and it is further away from point b by 0.90 meter. Also, the potential difference of an electric field is inversely proportional to the square of the distance from a point charge.

b. To calculate the magnitude (E) of the uniform electric field:

At a constant potential difference, the magnitude (E) of this electric field is given by the formula:

[tex]V_{ab} = Ed[/tex]

Where:

First of all, we would determine the distance (d).

[tex]d = d_b - d_a\\\\d = 0.90 -0.60[/tex]

Distance, d = 0.30 meter.

Substituting the given parameters into the formula, we have;

[tex]240 = 0.3E\\\\E=\frac{240}{0.3}[/tex]

E = 800 V/m

c. To calculate the work done on the point charge by the electric field:

Mathematically, the work done on a point charge in an electric field is given by the formula:

[tex]W = Fd = Eqd\\\\W = 800 \times (-0.200 \times 10^{-6}) \times 0.3\\\\W = -240 \times 0.200 \times 10^{-6}\\\\W = -48\times 10^{-6}\;Joules[/tex]

Note: 1 microjoules = [tex]1\times 10^{-6}[/tex] Joules

Work done = -48 microjoules.

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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:

Good conductor of heat

Explanation:

Because metals are shiny, ductile, malleable, sonorous, good conductors of heat and electricity and have high melting and boiling points

Answer: mine is different so im sorry im here for points

Explanation:

Answer:

power

Explanation:

Answer:

power

Explanation:

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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