Please help! This is due in 10 minutes

Answer:

it 1.5 meters

Explanation:

if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)

Answer:

t = 5.88 10⁻² s

Explanation:

The speed of the sound wave after it is emitted by the bat is constant, so we can use the uniform motion relationships

          v = [tex]\frac{x}{t}[/tex]

          t = [tex]\frac{x}{v}[/tex]

in this case the distance is that of the sound in going from the bat to the insect and back

         x = 2d

         x = 2 10

         x = 20 m

let's calculate

        t = 20/340

        t = 5.88 10⁻² s

We can see that the time is very short, so the distance traveled by the two animals has little influence on the result.

Answer:

     l = 0.548 m

Explanation:

For this exercise we compensate by finding the speed of the car

         p = m v

         v = p / m

         v = 0.58 / 0.2

         v = 2.9 m / s

this is how fast you get to the ramp, let's use conservation of energy

starting point. Lowest point

         Em₀ = K = ½ m v²

final point. Point where it stops on the ramp

         [tex]Em_{f}[/tex] = U = m g h

  mechanical energy is conserved

          Em₀ = Em_{f}

          ½ m v² = m g h

           h = [tex]\frac{m v^2}{2 g}[/tex]

let's calculate

          h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]

          h = 0.0858 m

to find the distance that e travels on the ramp let's use trigonometry, we look for the angle

          tan θ = y / x

          tan θ = 12/75 = 0.16

          θ = tan⁻¹ 0.16

          θ = 9º

therefore

           sin 9 = h / l

           l = h / sin 9

           l = 0.0858 / sin 9

           l = 0.548 m

Answer:

13.9 m/s.

Explanation:

Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.

So, distance = velocity time

h = vt

t = h/v = 70 m/7 m/s = 10 s

This is also the time it takes him to move horizontally a distance of d = 120 m to the target.

So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.

Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.

So, V = √(v² + v'²)

= √((7.0 m/s)² + (12.0 m/s)'²)

= √(49 m²/s² + 144 m²/s²)

= √(193 m²/s²)

= 13.9 m/s.

The direction of this velocity is Ф = tan⁻¹(v/v')

= tan⁻¹(7 m/s/12 m/s)

= tan⁻¹(0.5833)

= 30.3°

Answer:

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s

velocity of the train at t=12 s is;

[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s

now, we calculate the deceleration of the train

[tex]V_{t=12}[/tex]  = u + at

13.89 = 27.77 + [tex]a_{t}[/tex]12

[tex]a_{t}[/tex] = (13.89 - 27.77) / 12

[tex]a_{t}[/tex] = -13.88 / 12

[tex]a_{t}[/tex] = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

[tex]V_{t=6}[/tex]  = u + at

[tex]V_{t=6}[/tex]  = 27.77 + ( - 1.1566 m/s²)6

[tex]V_{t=6}[/tex]  = 27.77 - 6.9396

[tex]V_{t=6}[/tex]  = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

we substitute

2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]

4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²

4 = 1.3377 +  ([tex]a_{n}[/tex])²

([tex]a_{n}[/tex])² = 4 - 1.3377

([tex]a_{n}[/tex])² = 2.6623

[tex]a_{n}[/tex] = √2.6623

[tex]a_{n}[/tex]  = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² /  [tex]a_{n}[/tex]

[tex]p_{t=6}[/tex] = ( 20.83 m/s )² /  1.6316 m/s²

[tex]p_{t=6}[/tex] = 433.8889 / 1.6316

[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m

Therefore;  the radius of curvature of the track for this instant is 266 m

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  [tex]- \frac{dE}{dx} i^[/tex]

therefore the equipotential lines must also be perpendicular to the electric field lines

(a) The unit of electric potential is volts.

(b) Equipotential lines are lines along which the electric potential is constant.

(c) Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

Electric potential is the work done in moving a unit positive from infinite to a point in the electric filed.

Electric potential = EQ

Electric potential = (V/C) x (C)

Electric potential = Volts.

This is a line along which the electric potential is constant.

Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

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Answer:

(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C

Explanation:

The given parameters are;

The force of attraction between the two spheres = 0.0988 N

The distance between their centers = 44.5 cm = 0.445 m

Therefore, we have;

[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{d^2}[/tex]

Therefore, we have;

[tex]0.0988 \ N = -\dfrac{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}\cdot q_1 \cdot q_2}{(0.445 \ m)^2}[/tex]

Therefore, we have;

q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²

q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)

When the two charges are connected, we get;

[tex]F_2 = \dfrac{k \cdot \left (\dfrac{q_1 + q_2}{2} \right) ^2}{d^2}[/tex]

Therefore, we have;

[tex]q_1 + q_2 = \sqrt{\dfrac{4 \cdot F_2 \cdot d^2}{k} }[/tex]

[tex]q_1 + q_2 = \sqrt{\dfrac{4 \times 0.0276 \ N \times(0.445 \ m)^2}{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}} } = 1.59446902743 \times 10^{-6} \ C[/tex]

q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)

From, equation (2), we have;

q₁ = 1.59446902743 × 10⁻⁶ C -  q₂

Plugging in the value of q₁ in equation (1) givens;

q₁·q₂ = -2.17629255 × 10⁻¹²

Therefore, we have;

(1.59446902743 × 10⁻⁶ -  q₂) × q₂ = -2.17629255 × 10⁻¹²

Which gives;

-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0

Solving, with a graphing calculator, we get;

q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C

q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C

Therefore, we have;

(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C

Answer:

What was it

Explanation:

It would increase; g is inversely proportional to the square of the radius of the Earth. The correct option is A.

A current idea in geology that describes how the earth's crust is made up of a few big, hard plates that move independently of one another, causing deformation, volcanism, and seismic activity along their boundaries.

Because it explains how mountain ranges, earthquakes, volcanoes, shorelines, and other features often emerge where the moving plates contact along their boundaries, plate tectonics provides "the overall picture" of geology.

The Earth has been shrinking as it gradually cools, according to some geological hypotheses from the nineteenth century that have now been completely debunked.

If that were the case, it would rise since g is inversely proportional to the square of the Earth's radius.

Thus, the correct option is A.

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Answer:

Acceleration

Explanation:

Answer: Acceleration

Explanation:

Answer:

the answer is transmitted

Explanation:

Answer:

d = 29.89 m

Explanation:

To solve this, we need to separate this problem in two parts.

One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.

Joining these two times we have:

t = t₁ + t₂    (1)

This time is 2.56 s.

Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.

Doing this we have that the distance traveled by the rock is:

y₁ = gt²/2

y₁ = 9.8t²/2 = 4.9t₁²    (2)

Now, the distance traveled by sound would be:

y₂ = v * t₂ = 336t₂   (3)

Remember that the speed of the sound is 336 m/s

From this last expression (3), we can actually write t₂ in function of t₁, using (1):

2.56 = t₁ + t₂

t₂ = 2.56 - t₁    (4)

Replacing (4) in (3):

y₂ = 336(2.56 - t₁)    (5)

Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:

4.9t₁² = 336(2.56 - t₁)

4.9t₁² = 860.16 - 336t₁

4.9t₁² + 336t₁ - 860.16 = 0

Using the quadractic formula, we can calculate t₁:

t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)

t₁ = -336 ±√129,7555.136 / 9.8

t₁ = -336 ± 360.21 / 9.8     Using only the positive value we have:

t₁ = 2.47 s

This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)

With this, we can calculate the distance of the rock using expression (2):

y₁ = 4.9 * (2.47)²

Hope this helps

Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

Answer:

124.51 m

Explanation:

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

Initial velocity (u) = 49.4 m/s

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

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

The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:

v² = u² – 2gh ( since the ball is going against gravity)

0² = 49.4² – (2 × 9.8 × h)

0 = 2440.36 – 19.6h

Collect like terms

0 – 2440.36 = –19.6h

–2440.36 = –19.6h

Divide both side by –19.6

h = –2440.36 / –19.6

h = 124.51 m

Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m

Answer:

A

Explanation:

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = [tex]\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }[/tex]  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

a) determine the increase in speed

V = v - [tex]\sqrt{\frac{gR^2}{R + h} }[/tex]  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = [tex]\frac{3t}{2}[/tex]

 = [tex]\frac{3*5670.76}{2}[/tex]  =  8506.1 secs

attached below is the remaining part of the detailed solution

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Answer:

10 J.

Explanation:

Given that,

Net force acting on the rake, F = 2 N

Distance moved by the rake, d = 5 m

We need to find the kinetic energy gained by the rake. We know that,

Kinetic energy = work done

So,

K = F×d

K = 2 N × 5 m

K = 10 J

So, 10 J of kinetic energy is gained by the rake.

Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.

How much kinetic energy does the rake gain?

Answer: 10 J

Answer:

The ribbon will move to the right.

Explanation:

To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:

Force to the right (Fᵣ) = 120 N

Force to the left (Fₗ) = 80 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 120 – 80

Fₙ = 40 N to the right.

From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

Answer:

4

2

1

3

Explanation:

Be safe, lovelies <3

Answer:

Well, you said the answer is A, so it’s A!

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

[tex]W_{net}[/tex] = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

[tex]W_{net}[/tex] = m (gh)

m = [tex]W_{net}[/tex] / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

Answer:

B

Explanation:

Friction is a force that opposes motion between any surfaces that are touching. Friction occurs because no surface is perfectly smooth Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

Answer:

the time for the car to reach the final velocity is 0.56 s.

Explanation:

Given;

acceleration of the car, a = 50 m/s²

final velocity of the car, v = 100 km/h = 27.778 m/s

the initial velocity of the car, u = 0

The time for the car to reach the final velocity is calculated as;

v = u + at

27.778 = 0 + 50t

27.778 =  50t

t = 27.778 / 50

t = 0.56 s

Therefore, the time for the car to reach the final velocity is 0.56 s.

Answer:

energy= MGH

2*9.8*h=180

h=180/19.6

h=9.32 m

The height of the object above the ground would be equal to 9.18 m.

When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.

The mathematical equation for gravitational potential energy can be written as:

Gravitational potential energy = m⋅g⋅h

Where m is the mass, g is the gravitational acceleration and h is the height above the ground.

Given, the mass of the given object, m = 2 Kg

The gravitational potential energy = 180 J

[tex]GPE = m\times g\times h[/tex]

180 = 2 × 9.8 × h

h = 9.18 m

Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.

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Answer:

c

Explanation:

it is stored energy because it is built up in said object

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:

[tex]\tau = I\cdot \alpha[/tex] (1)

Where:

[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)

Where:

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega[/tex] - Final angular speed, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:

[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]

[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]

[tex]I = \frac{\tau}{\alpha}[/tex]

[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]

[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Answer:

convection currents

Explanation:

Answer:150M

Explanation: