I NEED THE KNOWLEDGE OF YOU SMART GUYS OUT THERE!!!!!!

I NEED THE KNOWLEDGE OF YOU SMART GUYS OUT THERE!!!!!!

Answers

Answer 1

                                    ...................


Related Questions

During the pitching motion, a baseball pitcher exerted an average horizontal force of 90 N against the 0.1 kg baseball while moving it through a horizontal displacement of 2.0 m before release. (1) what was the amount of work performed by the pitcher on the baseball (2) If the velocity at the start of the pitching motion was zero, at release the ball was traveling horizontally at which velocity

Answers

Answer:

(1) 180 J

(2) 60 m/s

Explanation:

(1) From the question,

Amount of work performed by the pitcher on the baseball = Force × distance.

W = F×d............... Equation 1

Given: F = 90 N,  and d = 2.0 m.

Substitute into equation 1

W = 90×2

W = 180 Joules.

(2)

F = ma........... Equation 2

Where a = acceleration of the ball.

a = F/m

Given: F = 90 N, m = 0.1 kg.

therefore,

a = 90/0.1

a = 900 m/s².

Using,

v² = u² + 2as ............ Equation 3

Where u = 0 m/s, a = 900 m/s², s  = 2 m.

substitute into equation 3

v² = 0² + 2×900×2

v² = 3600

v = √3600

v = 60 m/s

Find the quantity of heat needed
to melt 100g of ice at -10 °C
into water at 10 °C. (39900 J)
(Note: Specific heat of ice is
2100 Jkg 'K', specific heat
of water is 4200 Jkg K',
Latent heat of fusion of ice is
336000 Jkg ').​

Answers

Answer:

Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]

The energy required comes in three parts:

Energy required to raise the temperature of that [tex]0.100\; \rm kg[/tex] of ice from [tex](-10\; \rm ^\circ C)[/tex] to [tex]0\; \rm ^\circ C[/tex] (the melting point of ice.)Energy required to turn [tex]0.100\; \rm kg[/tex] of ice into water while temperature stayed constant.Energy required to raise the temperature of that newly-formed [tex]0.100\; \rm kg[/tex] of water from [tex]0\; \rm ^\circ C[/tex] to [tex]10\;\ rm ^\circ C[/tex].

The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:

[tex]Q = c \cdot m \cdot \Delta T[/tex],

where

[tex]c[/tex] is the specific heat capacity of the material,[tex]m[/tex] is the mass of the sample, and[tex]\Delta T[/tex] is the change in the temperature of this sample.

For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].

Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].

The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:

[tex]Q = m \cdot L_\text{f}[/tex].

Apply this equation to find the size of the second part of energy input:

[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].

Find the sum of these three parts of energy:

[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].

The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, given the period of one rotation is 3.15 x 107 s

Answers

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m =  5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

The linear momentum of earth is 17.8 * 10²⁸ kgm/s

The orbital radius (r) of earth is 1.496 x 10¹¹ m. Hence the distance covered by one rotation is:

Distance = 2πr = 2π(1.496 x 10¹¹) m

The period of one rotation is 3.15 x 10⁷ s.

The velocity of earth (v) = distance/time = 2π(1.496 x 10¹¹) m/ 3.15 x 10⁷ s = 298840 m/s

Linear momentum = mass * velocity = 5.972 x 10²⁴ kg * 298840 m/s = 17.8 * 10²⁸ kgm/s

Therefore the linear momentum of earth is 17.8 * 10²⁸ kgm/s

Find out more at: https://brainly.com/question/12194595

calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.

HELP PLSSS

Answers

Answer:

For the aceleration we have:

Vf = Vo + a * t

Clearing "a":

a = (Vf - Vo) / t

Replacing and resolving:

a = (34 m/s - 0 m/s) / 21 s

a = 34 m/s / 21 s

a = 1,61 m/s^2

The aceleration of the vehicle is 1,61 meters per second squared

A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:

Answers

Answer: Induced emf is given by:

ε= Bvl

On putting the values we get

=5×10

−5

×1.50×2

=0.15mV

Explanation:

Hope these helped have a wonderful Christmas break ❄️

A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?

Answers

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]

As there is a similar momentum

So,

[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power

Answers

P=350/300+20. P=350/320. P=35/32. P=1.09

Work done against gravity becomes potential energy of the object-Earth system. Its given by:

W = Ep

W = mgh  but weigh(w) = mg

W = wh

W = 350 N × 5 m

W = 1750 J

Power is the amount of energy transferred or converted per unit time. This is:

P = W/t

P = 1750 J / 20 s

P = 87.5 W

from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B

Answers

Answer:

1.8 m/s

Explanation:

Here is the complete question

The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.

What is the rate of discharge at point B?

Solution

Using Bernoulli's equation,

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)

(h₂ - h₁) = 5 m

Substituting the values of the variables, we have

700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)

36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)

- 13000 Pa = 500 kg/m³(v₂² - v₁²)

(v₂² - v₁²) = - 13000 Pa/500 kg/m³

(v₂² - v₁²) = -26 m²/s²

By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B

πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m

d₁²v₁ = d₂²v₂

v₁  = v₂(d₂/d₁)²

= v₂(0.15/0.05)²

= v₂(3)²

= 9v₂

So, substituting v₁ = 9v₂ into the above equation, we have

(v₂² - v₁²) = -26 m²/s²

v₂² - 9v₂² = -26 m²/s²

- 8v₂² = -26 m²/s²

v₂² = -26 m²/s² ÷ (-8)

v₂² = 3.25 m²/s²

taking square root of both sides,

v₂ = √(3.25 m²/s²)

= 1.8 m/s

So, the rate of discharge at point B is 1.8 m/s

Determine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equation: W or Fg = mg)

Answers

Answer:

15.3 sorry if i got it wrong

Explanation:

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Answers

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm[/tex]

Magnification,

[tex]m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25[/tex]

Hence, the magnification of the image is 0.25.

what is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher’s glove with a force of 25 newtons

Answers

Answer:

a = 50 [m/s²]

Explanation:

This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force =25 [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

[tex]a=F/m\\a=25/0.5\\a= 50 [m/s^{2}][/tex]

Which water would you use to make salt dissolve the slowest?

Boiling water
Cold water
Hot water
Room temperature water

Answers

Answer:

boiling water because salt dissolves quicker in hot Water and the hottest is boiling

A
B
с
4 teaspoons sugar
2 cups water
6 teaspoons sugar
2 cups water
4 teaspoons sugar
3 cups water
Which of these lemonade glasses is the most concentrated for sugar?
Ос

Oa

Answers

6 teaspoons of sugar

A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass

Answers

Answer:

let M be the mass of brick

let m be the mass of the pebble

.5M(3)^2 =KE

.5m(5)^2= KE

.5M9 = .5m25

9M = 25m

(9/25)M = m

Answer:

A

Explanation:

e=1/2M[tex]V^{2}[/tex]

1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25

[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]

If your parasympathetic nervous system was activated what might you be doing?

A. Relaxing under a tree
B. Running a marathon
C. Skiing
D. Driving home in traffic

Answers

A. Relaxing under a tree.

The parasympathetic nervous system is known as the “rest and digest” system. Activities that require as little movement as possible help the heart rate to slow down. The other activities would be associated with the Sympathetic nervous system which is known as the “fight or flight,” or when your heart rate is up and running.

The total yearly world consumption of energy is approximately 4.0 × 1020 J. How much mass would have to be completely converted into energy to provide this amount of energy?

Answers

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

[tex]m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}[/tex]

m = 4.4 × 10³ kg

A block has mass of 20 gm and volume of 4 m3 , calculate the density *

Answers

Answer:

Density = 5 g/m³

Explanation:

Given:

Mass of block = 20 gm

Volume of block = 4 m³

Find:

Density

Computation:

Density = Mass / Volume

Density = 20 / 4

Density = 5 g/m³

5 kg/m3

Explanation:

The formula to find density is (d) = mass (m) ÷ volume (v)

Why do we consider eyes, ears, and fingers as physical sensor​

Answers

Answer:

because

Explanation:

because we need eye to see if you don't have eyes how you regnise some one

if you have ears how can you know who's voice is who's

if don't have fingers how can you know what your are holding this is why they are called physical sensors

Answer:

because  they stimulate our brain and the sensor of human body detect stimuli.

Explanation:

A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare

Answers

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.

Answers

Answer:

330

Explanation:

(a) The torque the net thrust produces about the center of the circle is of 20 N-m.

(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

Given data:

The mass of model airplane is, m = 1.0 kg.

The radius of horizontal circle is, r = 20.0 m.

The magnitude of net thrust by engine is, F = 1.0 N.

(a)

The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,

T = F × r

Solving as,

T = 1.0 × 20.0

T = 20 N-m

Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.

(b)

The expression for the angular acceleration of airplane during the horizontal flight is given as,

[tex]T = I \times \alpha[/tex]

Here, I is the moment of inertia of airplane and its value is,

[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]

So, the angular acceleration is,

20 = 200 × α

α = 20/200

α = 0.1 rad/s²

Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

Learn more about the torque here:

https://brainly.com/question/19247046

9. Who was the FIRST person to propose that the continents might fit together

Answers

Answer:

Alfred Wegener

Explanation:

Correct answer:Alfred Wagner ! Hope this helps plz leave a thanks !

The arrow strikes a deer in the woods with the speed of 55 m/sec at an angle of 315 degrees. Calculate the Horizontal and vertical components of the arrow’s velocity.

Answers

Answer:

100 m

Explanation:

Helping me please What can smart modern biosensors be used in heath applications?​

Answers

Answer: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.

Explanation:

In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:

--> enzymatic biosensors

--> immuno biosensors

--> DNA and whole cell biosensors.

They can also be classified according to their signal transduction methods. These includes:

--> electrochemical Biosensors,

--> optical biosensors,

--> thermal and mass-based biosensors;

Some of the HEALTH APPLICATIONS of a modern biosensor includes:

--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.

--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.

--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.

--> They are used for Cancer clinical testings

Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select all that apply.
Heat energy
Geothermal energy
Tidal energy
Hydroelectric energy
Nuclear energy
Electrical energy
Solar Energy

Answers

Solar Energy is the answer to the question tell me if i`m wrong

If you are holding a stack of books at your waist and then you walk across the room your hands did?

Answers

Answer: 3) work

Explanation:

Work is doing something that requires you to expend energy. In walking across the room while your hands were carrying the stack of books, you expended energy through them which means that they did work.

We can neither tell the efficiency nor the power taken to do the work above from the given information. And by moving and holding books, there was definitely work done which means that all options apart from 3 are false.

A 3 Kg exercise ball is held 2m above the ground. What is the gravitational potential energy?​

Answers

Answer:

58.8

Explanation:

we should apply formula

m*g*h

3*9.81*2

how to calculate the speed using time and distance

Answers

Answer:

speed = distance/time

Explanation:

distance -> s

speed -> v

time -> t

The calculation of speed using time and distance shows that speed is equal to distance/time.

What is speed?

Speed refers to the rate of movement of a vehicle or person.

Thus, the speed can be computed by dividing the total distance by the time consumed.

Learn more about speed, distance, and time at https://brainly.com/question/4931057

a car accelerates from rest at 2 m/s. what is the speed after 8 sec?

Answers

Answer:

16m/s

Explanation:

[tex]v_{f}=v_{i}+at[/tex]

[tex]v_{f}=0+2\cdot8[/tex]

[tex]v_{f}=16\ \frac{m}{s}[/tex]

Therefore,  the speed after 8 seconds is 16m/s

Can someone please help?

Answers

Answer:

Gold = 19.3g/cm³

Explanation:

Given the following data;

Mass = 38.6g

Volume = 2cm³

To find the density;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

Substituting into the equation, we have;

[tex]Density = \frac{38.6}{2}[/tex]

Density = 19.3g/cm³

Therefore, from the table the material is Gold.

Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.

Generally, the chemical element Gold is known to have the following physical properties;

I. Malleable.

II. Ductile.

III. A good conductor of electricity and heat.

Also, it is a non-toxic chemical element with a beautiful lustrous sheen.

A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241

Answers

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

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