If you apply a net force of 3 N on 100g -box, what is the acceleration of
the box? *
30 m/s/s
20 m/s/s
10 m/s/s
5 m/s/s

Answer:

162

Explanation:

When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units. Therefore, option C is correct.

The term amplitude is defined as the maximum distance moved by a point on a vibrating body or wave calculated from its equilibrium position.

We have given,

first amplitude, A = 1, corresponding energy = 2

second amplitude, A = 2, corresponding energy = 8

The energy of a string at  a given amplitude is calculated as follows;

E ∝ A²

Energy varies as the square of the change in amplitude.

If amplitude of one has energy of two

Amplitude of 2 (double of 1)

= ( 2 x 2 ) = 4 x 2

= 8  4 times the energy at amplitude of 1.

Then for 8 units of amplitude change

8 x 8 = 64 x 2

= 128  

Thus, When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units, option C is correct.

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Your question is incomplete, most probably your question was

The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the mathematical relationship between amplitude and energy. Use your equation to find the energy if the amplitude is 8 units.

Amplitude

1

2

3

4

energy

2

8

18

32

A.66 units

B. 108 units

C. 128 units

D. 88 units ​

Answer:

the answer is A.) Wedge

A knife to cut a piece of bread is a A. Wedge

A simple machine, any of several devices with few or no moving parts that are used to modify motion and the magnitude of a force in order to perform work.

since , a wedge is a simple machine  with two inclined planes which when put together forms a sharped edge, which forms a triangular shaped tool which can be used to separate portion of  two objects .

hence , a knife to cut a piece of bread is a A. Wedge

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Answer:

Rating of percieved exertion

Answer:

Mechanical energy

Explanation:

Mechanical energy is needed for movement of objects. Muscles convert chemical energy provided by the rest of the body to allow movement.

Answer:

the bread is getting heated from top and bottom because the oven is heated all over and the particles from the oven makes the bread's particles heat up and the bread will be done

Explanation:

i have the same answer

Answer:

The air is heated by the heat source in the oven. The hot air rises, surrounds the bread, and heats the bread on all sides.

Explanation:

Answer:

8.8 hours

Explanation:

plz give me a Brainliest

Answer:

Explanation:

Time taken by jogger to travel the distance to finishing line = 2.1 / 7.3

= .28767 hr

Bird will keep flying for this time period

distance covered by bird = speed x time

= 29.2 x .28767 km

= 8.4 km .

Answer:

It's 31

Explanation:

add 15 and 16

Answer:

A. 50 km south

Explanation:

hope this helps!

Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  [tex]    R =  2281.6 \  cm  [/tex]

When mercury is used the reading is [tex]   R =  23.83 \ cm [/tex]  

The best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   [tex]P_1 = P_2[/tex]

Here [tex]P_1[/tex] is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       [tex] P_{atm} + P_1 =  P_{atm} + P_t[/tex]

Here [tex]P_t[/tex] is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         [tex]P_t  =  \rho_t  * g  * h_i[/tex]

Here  

      [tex]\rho_t [/tex] is the density of toluene with value  [tex]\rho_t =  867 kg/m^3 [/tex]

 [tex]h_i[/tex] is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          [tex]h_i =  d - h + R[/tex]

  and  [tex] P_2 [/tex] is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         [tex] P_2 =  \rho_f * g *  h_f [/tex]

Here  

       [tex]\rho_f[/tex] is the density of the fluid in use , if it is water the density is  

       [tex]\rho_w =  1000 \  kg /m^3 [/tex]

and  if it is  mercury the density is  

        [tex]\rho_m =  13600 \  kg /m^3 [/tex]

[tex]h_f[/tex] is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      [tex] P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f[/tex]

=>   [tex]   \rho_t* (d - h + R) =   \rho_w *  h_f[/tex]        

=>   [tex]    867  (500 - 150 + R) =    1000 *  R [/tex]

=>    [tex]    R =  2281.6 \  cm  [/tex]

So when the fluid is mercury we have      

  [tex]   \rho_t* (d - h + R) =   \rho_m *  h_f[/tex]          

=>   [tex]    867  (500 - 150 + R) =   13600  *  R [/tex]  

=>   [tex]   R =  23.83 \ cm [/tex]  

The difference in the mercury reading for mercury due to the fact that they have different densities as we have seen in this calculation

So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Answer:

D

Explanation:

I could be wrong, But I'm pretty sure it's D

Answer:

300m per minute or 5m per second

d) ≈ 0.87 m/s²

Hi there !

F = m×a => a = F/m

a = 59N/68kg

1 N = 1kg·m/s²

= (59kg·m/s²)/68kg

= 0.8676 m/s²

≈ 0.87 m/s²

Good luck !

Gravitational force between two cars are 2.92 × 10 ⁻¹¹ Nm²/kg.

The universal force of attraction happened between two objects is called as gravitational force.

The mass of the two cars, m₁ = 2565 kg

                                            m₂ = 4264 kg

                                             r = 5m

                                            G = 6.67 ₓ 10 ⁻11 Nm²/kg

How gravitational force can be calculated?

Definition of gravitational force: The force of attraction between any two bodies is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

                            F ∝ m₁ m₂ ÷ r²

where the force exerted between the two masses m₁ and m₂,

                            F = G (m₁ m₂) ÷ r²

where as,  F - Gravitational force between two bodies,

                 G - Gravitational constant,

                 m₁ - mass of the first body,

                m₂ - mass of the second body,

                r² - square of the distance between two bodies.

                        F= 6.67 × 10⁻¹¹ Nm²/kg ( 2565 kg × 4264 kg) ÷ ( 5²)

         Hence,

                         F = 2.92 × 10⁻⁵ N

Option D is correct answer.

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Answer:

UR ANSWER IS WAVELENGTH

Answer: Approximately 302 m/s^2

================================================

Work Shown:

s = starting velocity = 0

f = final velocity  = 46

d = distance = 3.5

a = acceleration = unknown (we're solving for this)

[tex]f^2 = s^2 + 2a*d \ \ \text{ ..... one of the kinematics equations}\\\\46^2 = 0^2 + 2a*3.5\\\\2116 = 7a\\\\7a = 2116\\\\a = \frac{2116}{7}\\\\a \approx 302.28571\\\\a \approx 302[/tex]

The acceleration to three sig figs is roughly 302 m/s^2

The acceleration is so large because the ball's final velocity is incredibly fast in such a short amount of time.

Answer:

Explanation:

The force acting on the object can be found by using the formula

[tex]f = \frac{w}{d} \\ [/tex]

where

d is the distance

w is the work done

We have

[tex]f = \frac{3600}{45} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater ...

Answer:

a.) 10Hz

b.) 0.1 s

c.) 187.4 m/s

d.) -412.6 m/s^2

Explanation:

Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.

(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?

from the equation given,  the angular speed w = 20π

but w = 2πf

where f = frequency.

substitute w for 20π

20π = 2πf

f = 20π/2π

f = 10 Hz

(b) What is t the first time that the object is at its farthest right?

since F = 1/T

T = 1 / f

T = 1/10

T = 0.1 s

Therefore, the t of  first time that the object is at its farthest right is 0.1 s

(c) At the time found in part (b), what is the object's velocity?

The velocity can be found by differentiating the equation;

x(t) = 3sin(20πt)

dx/dt = 60πcos(20πt)

where dx/dt  = velocity V

V = 60πcos(20π * 0.1)

V = 187.4 m/s

(d) At the time found in part (b), what is the object's acceleration?

to get the acceleration, differentiate equation  V = 60πcos(20πt)

dv/dt = -1200πSin(20πt)

dv/dt = acceleration a

a = -1200πSin(20πt)

substitute t into the equation

a = -1200πSin(20π * 0.1)

a = - 412.6 m/s^2

The type of motion the object undergoes that consist of a back-and-forth

movement that is repetitive is a simple harmonic motion.

The correct responses are;

Reasons:

The equation that represents the motion of the object is x(t) = 3·sin(20·π·t)

Where;

t ≥ 0

x(t) is in cm, and t is in seconds.

(a) The general sine function is y = A·sin(B·(t - C) + D

[tex]\mathrm{The \ period \ P} = \dfrac{2 \cdot \pi}{B}[/tex]

By comparison, we have;

B = 20·π

Therefore;

[tex]P = \dfrac{2 \cdot \pi}{20 \cdot \pi} = \dfrac{1}{10} = 0.1[/tex]

The period, which is the time to complete one cycle = 0.1 seconds

The number of cycle completed per second, is the frequency, f

[tex]f = \dfrac{1}{P} = \dfrac{1}{0.1} = 10[/tex]

f = 10 Hz = 10 cycles

The number of cycle completed per second is f = 10 cycles

(b) When the object is at the farthest right, we have;

sin(20·π·t) = Maximum = 1

[tex]\mathbf{sin\left(\dfrac{\pi}{2}\right)} = 1[/tex]

Therefore,

[tex]20 \cdot \pi \cdot t = \dfrac{\pi}{2}[/tex]

[tex]t = \dfrac{\pi}{20 \cdot \pi \cdot 2} = \dfrac{1}{40} = 0.025[/tex]

The first time it is at the farthest right is t = 0.025 seconds after start

(c) [tex]\mathrm{Velocity, \ v, \ is \ given \ by \ v = \dfrac{dx(t)}{dt}}[/tex]

Therefore;

[tex]v = \dfrac{dx(t)}{dt} = \dfrac{dx(t)}{dt} \left(3 \cdot sin \left(20 \cdot \pi \cdot t \right) = 60\cdot \pi \cdot cos \left(20 \cdot \pi \cdot t \right)[/tex]

At t = 0.25 seconds, we have;

[tex]v = \dfrac{dx(t)}{dt} = 60\cdot \pi \cdot cos \left(20 \times \pi \times 0.025 \right) = 0[/tex]

The velocity of the object at the t = 0.025 seconds is v(t) = 0

(d) Acceleration, a, is given by [tex]a = \dfrac{dv(t)}{dt}[/tex]

Therefore;

[tex]a = \dfrac{dv(t)}{dt} = \dfrac{d}{dt} \left( 60\cdot \pi \cdot cos \left(20 \times \pi \times t\right) \right) =-1200 \cdot \pi ^2 \cdot sin(20 \cdot t \cdot \pi)[/tex]

a = -1200·π²·sin(20·t·π)

∴ a = -1200×π²×sin(20×0.025×π)

sin(20×0.025×π) = 1

∴ a = -1200×π²×1 ≈ -11843.53

At the time found in part b the acceleration of the object, a ≈ -11843.53 m/s²

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To make this problem solvable and you can get the help you need, I'll complete and arrange some data.

Answer:

Acceleration: [tex]0.0417\ m/s^2[/tex], Distance=7,500 m

Explanation:

Uniform Acceleration Motion

It's a type of motion in which the velocity of an object changes uniformly over time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

[tex]v_f=v_o+at\qquad\qquad [1][/tex]

The distance traveled by the object is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]

Using the equation [1] we can solve for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}\qquad\qquad [3][/tex]

The problem will be rewritten as follows:

A train starting from rest reaches a velocity of 90 km/h in 10 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by the train for attending the velocity​.

Let's take the relevant data:

vo=0

vf=90 Km/h*1000/3600 = 25 m/s

t = 10 minutes = 10*60 = 600  seconds

Now compute the acceleration by using [3]:

[tex]\displaystyle a=\frac{25-0}{600}=0.0417[/tex]

[tex]a=0.0417\ m/s^2[/tex]

Finally, compute the distance:

[tex]\displaystyle x=0*600+\frac{0.0417\cdot 600^2}{2}[/tex]

[tex]x=7,500\ m[/tex]

Note: We used the value of the acceleration with more precision than shown.

Acceleration: [tex]\mathbf{0.0417\ m/s^2}[/tex], Distance=7,500 m

Answer:

The correct option is the third option

Explanation:

Firstly, it must be noted that chemicals/reagents that are sensitive to sunlight are stored in dark/amber container in the laboratory. Hence, the chloroform can only be found in an amber/dark bottle.

Also, reagents/chemicals that release poisonous/offensive gases are handled in the fume cupboard in the laboratory. Thus, If Malik is going to pour the chloroform, he should pour it in a fume cupboard to avoid inhaling it because of the toxicity of it's vapor.

From the above explanation, it can be deduced that Malik should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Answer:

He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Explanation:

Answer:

This has 5 significant figures.

Answer: The process is called artificial selection because people (instead of nature) select which organisms get to reproduce.

Explanation: This is evolution through artificial selection and the characteristics to reproduce, causing the evolution of farm stock.

Answer:

9,200 m

Explanation:

1 kilometer = 1000 meters so if there is 9 kilometers then it is 9000 meters, then .2 kilometer is equal to 200 meters. Add the numbers together to get 9,200 meters

Answer:

on earth yes, but in space no because there is no gravity

Explanation:

plz can i have some points

The force exerted by an object depends on the mass of the object. Greater mass results in greater force. Hence, the statement is true.

Force is an external agent acts on a body to change the substance from the state of rest or motion. There are various kinds of force such as nuclear force, magnetic force, frictional force etc.

According to Newton's law of motion force in a moving body is the product of its mass and acceleration. Hence, force is directly proportional to mass and acceleration and an increase in these quantities will increase the force exerted by the body.

Thus, greater force is needed to a move a body with greater mass. If two objects of different masses exerts force each other, the greater force will be exerted by the massive body.

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Explanation:

Acceleration is the rate of change in the speed of an object. If initial and final speed are v₁ and v₂. It changes speed in time t. Its formula is given by :

[tex]a=\dfrac{v_2-v_1}{t}[/tex]

The problem is "A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the skater’s acceleration? "

In this problem,

v₁ = 2 , v₂ = 10 m/s and t = 3 s

Acceleration of the object,

[tex]a=\dfrac{10-2}{3}\\\\a=2.67\ m/s^2[/tex]

So, the acceleration of the skater is [tex]2.67\ m/s^2[/tex].

Answer: Measurement is the assignment of a number to a characteristic of an object or event, which can be compared with other objects or events. The scope and application of measurement are dependent on the context and discipline. In the natural sciences and engineering, measurements do not apply to nominal properties of objects or events, which is consistent with the guidelines of the International vocabulary of metrology published by the International Bureau of Weights and Measures. However, in other fields such as statistics as well as the social and behavioural sciences, measurements can have multiple levels, which would include nominal, ordinal, interval and ratio scales.

Explanation:

Answer:

The correct option is;

C. The temperature of the air affects the speed of sound

Explanation:

The information given are;

The speed with which sound travels in the desert air = 358 m/s

The speed with which sound travels in the polar air = 330 m/s

The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.

The equation that gives the relationship of the speed of sound in air, [tex]v_{air}[/tex], to temperature is presented as follows;

[tex]v_{air} = 331\frac{m}{s} \times \sqrt{\dfrac{T_K}{273 K} } = 331\frac{m}{s} \times \sqrt{1 + \dfrac{T_{^{\circ}C}}{273 ^{\circ} C} }[/tex]

Which shows that the speed of sound in air, [tex]v_{air}[/tex], rises as the temperature of the air rises.

Where;

[tex]T_{^{\circ}C}[/tex] = The temperature of the air in degrees Celsius

[tex]T_K[/tex] = The temperature of the air in degrees Kelvin.