Answer:
The answer is "91 m"
Explanation:
[tex]v_0 = 2.5 \ \frac{m}{s}\\\\a = 4.2\ \frac{m}{s^2}\\\\t = 6.0\ s\\\\\Delta x=?[/tex]
Using formula:
[tex]\Delta x = v_0 t +\frac{1}{2} at^2\\\\[/tex]
[tex]= 2.5 \times 6.0 +\frac{1}{2} \times 4.2 \times 6.0^2\\\\= 90.6\ m \approx 91\ m[/tex]
Pls help me with this fast. I will mark brainiest
Answer:
a) 70, 95
b) 95-70= 25cc
c) density= mass/volume
102/25
=4.08g/cc
A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g
(1 -cos²)/ cos θ = [tex]\frac{v^2 }{g \ L}[/tex]
1 - cos² θ = [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex] cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]
T cos θ = m g
resolved
tan θ = [tex]\frac{v^2}{ r g}[/tex]
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
Calculate the efficiency of the following appliances:
1. A radiator that converts 1000) of electrical energy into 900J
of heat energy and 100J of light energy
2. A torch that converts 100J of chemical energy into 35) of
light energy and 65J of heat energy
3. A car that converts 10,000J of chemical energy into 6000) of
kinetic energy and 4000J of heat energy.
4. An energy saver light converts 1,000J of electrical energy
into 7003 of light energy and 300J of heat energy.
5. A speaker converts 100J of electrical energy into 50J of
sound energy and 50J of heat and kinetic energy.
we
Answer:
1. The efficiency of the radiator is 90 %
2. The efficiency of the torch is 65 %
3. The efficiency of the car is 40 %
4. The efficiency of the energy saver is 70 %
5. The efficiency of the speaker is 50 %
Explanation:
Efficiency = (Useful energy out ÷ Total energy in) × 100 J
1. Useful energy = 900 J
The total energy in = 1000 J
The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %
2. Useful energy = 65 J
The total energy in = 100 J
The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %
3. Useful energy = 4,000 J
The total energy in = 10,000 J
The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %
4. Useful energy = 700 J
The total energy in = 1,000 J
The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %
5. Useful energy = 50 J
The total energy in = 100 J
The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %
The capacity of an RD bottle is 100 cc and its mass is 45 g. Find the mass of the bottle when it is filled with a liquid of density 600 kg/m3.
Answer:
The capacity of an RD bottle is 100cc and its mass is 45g. Find the mass of the bottle when it is filled with a liquid of density 600kg/[tex]m^{3}[/tex]
EXPLANATION:
100cc=100[tex]cm^{3}[/tex]
density of liquid=600Kg/ =0.6gm/
so mass of 100cc liquid will be m=d×V=0.6×100=60gm
So mass of bottle filled with liquid will be bottle+liquid=45gm+60gm=105gm
Answer is 105gm.
define one standard kg
Answer:
The mass of platinum iridium rod whose diameter and height are equal an kept at international bureau of weight and measurement in Paris of France is called as one standard kg.
The unit of work in terms of Newton and metre is _______
Answer:
Joule
Explanation:
the unit of work is Joule determined by the letter J.
Answer:
JouleExplanation:
Joule is defined as the work done by a force of one newton causing a displacement of one meter. Sometimes, newton-metre (N-m) is also used for measuring work.
: 1 khối khi lý tưởng nhận được nhiệt lượng 200J, khi đó khí nở ra đẩy pittong bằng 1 công 80J.
Vậy nội năng Của khí là bao nhiêu
Answer:
DU = 120 Joules
Explanation:
Given the following data;
Quantity of energy = 200 J
Work = 80 J
To find the change in internal energy;
Mathematically, the change in internal energy of a system is given by the formula;
DU = Q - W
Where;
DU is the change in internal energy.
Q is the quantity of energy.
W is the work done.
Substituting into the formula, we have;
DU = 200 - 80
DU = 120 Joules
there is a dropping of a coconut of 2kg mass. the speed of the coconut at the moment before it hits the ground is 40m s-1.
calculate the kinetic energy of the coconut at the moment.
We know
[tex]\boxed{\sf K.E=\dfrac{1}{2}mv^2}[/tex]
m denotes to mass v denotes to velocity[tex]\\ \sf\longmapsto K.E=\dfrac{1}{2}\times 2\times 40^2[/tex]
[tex]\\ \sf\longmapsto K.E=40^2[/tex]
[tex]\\ \sf\longmapsto K.E=1600J[/tex]
what is the role of the communicator in a dc generator?
A) it keeps the electric current flowing in one direction
B) it keeps the magnetic fields of the permanent magnet steady
C) it causes the magnetic fields of the permanent magnets to reverse
D) it causes the electric current to constantly reverse its direction
Answer:
C
Explanation:
Answer:
A)
Explanation:
it keeps the electric current flowing in one direction
what you filling your heart with
oxygen and blood
Answer:
Explanation:
The right side of your heart receives oxygen-poor blood from your veins and pumps it to your lungs, where it picks up oxygen and gets rid of carbon dioxide. The left side of your heart receives oxygen-rich blood from your lungs and pumps it through your arteries to the rest of your body.
#I AM ILLITERATE
Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=4.00×103 kg and the second a mass of m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?
Answer:
Their final relative velocity is 0.190 m/s
Explanation:
The relative velocity of the satellites, v = 0.190 m/s
The mass of the first satellite, m₁ = 4.00 × 10³ kg
The mass of the second satellite, m₂ = 7.50 × 10³ kg
Given that the satellites have elastic collision, we have;
[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]
[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]
Given that the initial velocities are equal in magnitude, we have;
u₁ = u₂ = v/2
u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s
v₁ and v₂ = The final velocities of the satellites
We get;
[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂
∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190
The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s
Answer me as much as you can
Answer:
b) Fluorescent powder.
4 A student says that he has made a magnetic field with some iron filings. What would
you say to him?
Answer:
Yes but no
Explanation:
Here the kid cant say that he has created s magnetic field as iron fillings do not have magnetic properties
What is an effect of continental drift?
Answer: An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.
an athlete had lifts a load with a mass of 150kg.
1) calculate the gravitational potential energy gained
2) if the mass of the load is increased to 200kg, calculate the gravitational potential energy gained by the load.
3) based on answers in 1&2, state the relationship between the mass of the load and the gravitational potential energy.
[tex]\\ \sf\longmapsto P.E=mgh[/tex]
[tex]\\ \sf\longmapsto P.E=150(2)(10)[/tex]
[tex]\\ \sf\longmapsto P.E=3000J[/tex]
In 2nd case
Mass =m=200kgHeight=h=2mg=10m/s^2[tex]\\ \sf\longmapsto P.E=200(2)(10)[/tex]
[tex]\\ \sf\longmapsto P.E=4000J[/tex]
We can observe that
If mass of body increases gravitational potential energy will increase.Define SI units. Also mention their importance. SI unit The
Answer:
SI unit is an internationally accepted system and have the same value all over the world .
I don't know the importance sotry
Answer:
SI unit is an internationally accepted system that have same value all over the world.
Which of the following processes occurs in a battery?
Batteries convert mechanical energy into chemical energy.
Batteries convert chemical energy into kinetic energy.
Batteries convert mechanical energy into electric energy.
Batteries convert chemical energy into electric energy.
Answer:
Batteries convert chemical energy into electric energy.Explanation:
As seen, batteries have chemical energy in them stored which when used in any electronics produces electricity.
Thus, we can conclude that a battery converts chemical energy into electric energy.
The rate of chage of displacement is called __________
Answer:
Velocity is the rate of change of displacement.
the velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration
Answer:
acceleration = v-u/ t
= 15-10/5
= 5/5
= 1 m/s2
Explanation:
hope this helped you.
Answer:
[tex]\boxed {\boxed {\sf 1 \ m/s^2}}[/tex]
Explanation:
Acceleration is the change in velocity over the change in time. Therefore, the formula for calculating acceleration is:
[tex]a= \frac{v_f-v_i}{t}[/tex]
Since the body's velocity increased from 10 meters per second to 15 meters per second 15 m/s is the final velocity and 10 m/s is the initial velocity. The time is 5 seconds.
[tex]v_f[/tex]= 15 m/s [tex]v_i[/tex]= 10 m/s t= 5 s[tex]a= \frac{ 15 \ m/s - 10 \ m/s}{5 \ s}[/tex]
Solve the numerator.
15 m/s - 10 m/s = 5 m/s[tex]a= \frac{ 5 \ m/s }{5 \ s}[/tex]
Divide.
[tex]a= 1 \ m/s/s[/tex]
[tex]a= 1 m/s^2[/tex]
The acceleration is 1 meter per second squared.
Why is velocity proportial to pressure?
Answer:
Pressure and velocity are inversely proportional to each other because if pressure increase, the velocity decrease to keep the algebraic sum of potential energy, kinetic energy and pressure constant.
What is 98.907 rounded to 1 significant figure?
The answer to the digits rounded to one significant figure is 100.
Significant figures are the most important figures in a number of digits.
The closer a digit is to the beginning of a number, the more important or significant it is.
But, the rule for rounding up a set of numbers to the first significant or to one significant figure is that:
If the next number is 5 or more, we round up.If the next number is 4 or less, we do not round up.From the numbers 98.907, 9 is the most significant digit followed by 8.
8 is greater than 5 so we should round up by adding one to 9 making the number 100.
learn more here:
https://brainly.com/question/11566364
The coefficient of kinetic friction between an 100-kg desk and the wood floor is 0.4. What force (in Newtons) must be applied to move the desk at a constant speed? help now
Assuming the applied force is exerted parallel to the floor, by Newton's second law both the net vertical and net horizontal forces would be zero:
∑ F (horizontal) = p - f = 0 … … … because the desk is pushed at a constant speed
∑ F (vertical) = n - mg = 0 … … … because the desk doesn't move up or down
where p is the magnitude of the applied force, f is the mag. of kinetic friction, n is the mag. of the normal force due to contact between the floor and desk, and mg is the weight of the desk. We have
n = mg = (100 kg) g = 980 N
and the mag. of friction is proportional to n according to
f = 0.4 n = 392 N
Then the applied force p has magnitude
p = f = 392 N
define inertia.mention it's types
Answer:
It is the inability of the body to change by itself its state of rest or uniform motion or direction. Types of Inertia- It is of three types-(1)Inertia of rest (2) Inertia of motion(3) Inertia of direction. (1) Inertia of rest - It is the inability of the body to change by itself its state if rest.
3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?
Answer:
Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.
Please help me. I need this in a few hours.
Answer:
A) The population of prey is greater than that of predator
A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by the mirror?
• -5.0 cm
• 7.5 cm
• -7.5 cm
• 5.0 cm
Answer:
5.0cm
Explanation:
To get the height of the image, we will use the magnification formula as shown:
m = Hi/H = v/u
Hi is the image height
H is the object height
v is the image distance
u is the object distance
First, we need to get the image distance v
Using the mirror formula;
1/f = 1/v+ 1/u
1/-6 = 1/v + 1/12
1/v = -1/12 - 1/6
1/v = -1-2/12
1/v = -3/12
v = 12/-3cm
v = -4cm
Next is to get the image height Hi
Using the expression;
Hi/H = v/u
Hi/15 = 4/12
Hi/15 = 1/3
3Hi = 15
Hi = 15/3
Hi = 5.0cm
Hence the image height is 5.0cm
Answer:
D. 5.0 cm
Explanation:
got it correct on the test
what is measurement?
Answer:
The comparison of unknown quantity with known quantity is called measurement
Answer:
the size, length or amount of something is a measurement.
Explanation:
Measurements help in daily life and are important to development, accuracy and making decisions.
300 ml of a gas at 27°C is Cooled at -3°c at Constant pressure the final volume is plzz answer fast i will mark brainliest
Answer:V₁=300ml
T₁=27°C
V₂=?
T₂= -3°C
as we know
V₁T₁=V₂T₂
By putting values in formula
300ml×27°C=V₂×(-3°C)
300ml×27°C/-3°C=V₂
8100ml/-3=V₂
-2700ml=V₂
or V₂= -2700ml
010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave base is the lowest sea surface elevation attained during the passage of a wave. Waves have no effect on the seafloor near the shoreline. Waves speed up as they enter shallower water.'
Answer:
The height of the wave is determined by the wind strength and fetch.
Explanation:
The height of the wave is determined by the wind strength and fetch.
The more the strength and the more the fetch size the more will be the height of the wave.
Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.
maize is a monocotyledonous seed and pea is a dicotyledonous seed why? give short and the suitable answer I will mark you as a brainelist
Answer:
A dicot is a flowering plant that has one seed leaves. The monocot plants have a single cotyledon. Maize only has one cotyledon in their seed, so it's a monocot. Seeds having two Cotyles are mainly called a Dicot. A pea is a dicotyledonous plant, the seed (the pea itself) has two halves, cotyledons, hence dicot being 2.
Explanation:
One or more of the cotyledons are the first to appear from a germinating seed. Based on the number of cotyledons, botanists classify flowering plants (angiosperms) into :
a) plants with one embryonic leaf, termed monocotyledonous (monocots).
b) plants with two embryonic leaves, termed dicotyledonous (dicots).
Helpful Link:
https://www.vedantu.com/question-answer/in-pea-caster-and-maize-the-number-of-cotyledons-class-11-biology-cbse-5f626a17e5bde9062ff6d2a3