Answer:
a). Silver, Ag
b). Iron, Fe
c). Copper, Cu
d). Aluminium, Al
e). Gold, Au
round off 20.96 to 3 significant figures. a.20.9 b.20 c.21.0 d.21
Answer:
option c. 21.0
Explanation:
It was given that to find 3 significant figures. So the answer is 21.0
10. Match the following varibles to their relationship in Newton's 2nd Law. Questions 1. Force and Acceleration 2. Mass and Acceleration 3. Speed and Distance Answer Choices A. Direct Relationship B. Inverse Relationship C. Not in Newton's 2nd Law
Explanation:
based on the above information
1.A
2.B
3. C
1. A bicycle initially moving with a velocity
5.0 m s-1 accelerates for 5 s at a rate of 2 m s? Wh
will be its final velocity ?
Answer:
[tex]\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}[/tex]
Explanation:
We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.
[tex]v_f= v_i+ at[/tex]
In this formula, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, and [tex]t[/tex] is the time.
The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.
[tex]\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)[/tex]
Solve inside the parentheses.
[tex]\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s[/tex][tex]v_f= 5.0 \ m/s + (10 \ m/s)[/tex]
Add.
[tex]v_f= 15 \ m/s[/tex]
The units can also be written as:
[tex]v_f= 15 \ m*s^{-1}[/tex]
The bicycle's final velocity is 15 meters per second.
A comet of mass 2 × 10^8 kg is pulled toward the star. If the comet's initial velocity is very small, and the comet starts moving toward the star from 700,000,000 km away, how fast is it going right before it hits the surface of the star? (Assume that it does not lose any mass by melting as it approaches the star.)
Answer:
The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s
Explanation:
Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg
Explanation;
The given mass of the comet, m = 2 × 10⁸ kg
The initial velocity of the comet, v → 0
The distance of the comet from the star, d = 700,000,000 km
The gravitational potential at d = G·M·m/d
The kinetic energy of the comet, K.E. = m·v²/2
The kinetic energy of the comet at d = m·(0)²/2 = 0
The gravitational potential at the surface of the star, R = G·M·m/R
The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0
Where;
M = The mass of the star = 5 × 10³¹ kg
[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg
M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
R = The radius of the star
Therefore, we have;
m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R
∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))
Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))
v = 81696389149.1×√(1/R - 1/700,000,000,000).
The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)
The mass radius relationship is given as follows;
[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]
[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]
The radius of the Sun = 696,340,000 M
∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76
R = 4538865694.76 m
v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7 m/s
Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed
Answer:
condensation - thermal energy removed
freezing -thermal energy removed
deposition - thermal energy removed
sublimation - thermal energy added
evaporation - thermal energy added
melting - thermal energy added
Explanation:
Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.
Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.
On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.
Answer:
condensation - thermal energy removed
freezing -thermal energy removed
deposition - thermal energy removed
sublimation - thermal energy added
evaporation - thermal energy added
melting - thermal energy added
Que. I : A mass of 10kg is suspended from the end of a steel of length 2m and radius 1mm, what is the elongation of the rod beyond its original length?
Que 2 : A pressure of sea water increases by 1.0atm for each 10metres increase in the depth. by what what percentage is the density of water increased in the deepest ocean of about 12km; compressibility = 5.0 × 10^-5
Question 1; The elongation of the steel is approximately 0.3123 mm
Question 2; The percentage the density of water increased in the deepest
ocean is approximately 6.4%
The strategy of obtaining the above solution is presented as follows;
Que. 1; The given parameters are;
The mass of the suspended block, m = 10 kg
The length of the steel, l = 2 m
The radius of the steel, r = 1 mm = 1 × 10⁻³ m
The modulus of elasticity of steel, E = 200 GPa = 200 × 10⁹ Pa
The stress, σ, on the steel due to the mass, m, is given as follows;
[tex]\mathbf{\sigma = \dfrac{F}{A}}[/tex]
Where;
F = The force acting on the steel = The weight of the mass
A = The cross sectional area of the steel = π·r²
∴ F = 10 kg × 9.81 m/s² = 98.1 N
A = π × (1 × 10⁻³)² = 3.14159 × 10⁻⁶ m²
Therefore;
σ = 98.1 N/(3.14159 × 10⁻⁶ m²) ≈ 31,226,226.2 Pa
We have;
[tex]\mathbf{ E = \dfrac{\sigma}{\epsilon}}[/tex]
From which we have;
[tex]\epsilon = \dfrac{\sigma}{E}[/tex]
Where;
∈ = The tensile strain = Δl/l
Δl = The elongation of the steel
Therefore;
∈ = 31,226,226.2/(200 × 10^9) = 0.00015613113
∴ Δl = 0.00015613113 × 2 m = 0.00031226226 m = 0.31226226 mm
The elongation of the steel, Δl = 0.31226226 mm ≈ 0.3123 mm
Question 2
The given parameters are;
The change in pressure per unit depth, Δp = 1.0 atm per 10 meters
The depth of the ocean = 12 km = 12,000 m
The compressibility = 5.0 × 10⁻⁵
The formula for compressibility, C, is presented as follows;
[tex]C = \dfrac{1}{V} \times \dfrac{\partial V}{\partial P}[/tex]
The change in pressure, [tex]\partial P[/tex] = 12,000 m × 1.0 atm/(10 m) = 1,200 atm
For a unit volume, V = 1 m³
We get;
[tex]5 \times 10^{-5} = \dfrac{1}{1} \times \dfrac{\partial V}{1,200}[/tex]
[tex]\partial V[/tex] = 5 × 10⁻⁵ m³/(atm) × 1,200 = 0.06 m³
The volume occupied 1 m³ at 12,000 km depth = V - [tex]\partial V[/tex]
∴ The volume occupied 1 m³ at 12,000 km depth = 1 m³ - 0.06 m³ = 0.94 m³
The percentage density increase, [tex]\partial[/tex]ρ% = (m/0.94 - m/1)/m/1 × 100
∴ (1/0.94 - 1/1)/1/1 × 100 ≈ 6.4%
The percentage increase in density ≈ 6.4%
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If Earth's gravity pulls an object, causing it to accelerate to the ground, what
must be true about Earth?
A. It accelerates just as quickly in the direction away from the object.
B. It is being pulled toward the object by the object's gravity.
C. It accelerates just as quickly in the direction of the object.
D. It is being pushed away from the object by that same force.
Ans
It is being pulled toward the object by the objects gravity
Gravity:-
Sir Eizak Newton Founded the gravity.Gravity is a force between any object and earth by which they pull each other .The Acceleration due to gravity is represented by g =9.8m/s^2I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work
This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.
The acceleration of the bird is "a = 26.13 m/s²".
First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:
[tex]h = v_it + \frac{1}{2}gt^2[/tex]
where,
h = height fall = 3 m
vi = initial velocity = 0 m/s
g = acceleration due to gravity = 9.8 m/s²
t = time taken = ?
Therefore,
[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]
The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:
[tex]s = v_it + \frac{1}{2}at^2[/tex]
where,
s = distance covered by the bird = 5 m + 3 m = 8 m
vi = initial velocity of the bird = 0 m/s
a = acceleration of the bird = ?
t = time taken = 0.78 s
Therefore, using these values we get:
[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]
a = 26.13 m/s²
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Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường?
Answer:
Distance = 6.667 kilometres
Explanation:
Given the following data;
Speed = 20 km/h
Departure time = 7:00
Arrival time = 7:20
Time taken = 20 minutes
To calculate the distance travelled from home to school;
First of all, we would have to convert the value of time in minutes to hours.
Conversion:
60 minutes = 1 hour
20 minutes = X hours
Cross-multiplying, we have;
X = 20/60 = 1/3 hours
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 20 * 1/3
Distance = 20/3 =
Distance = 6.667 kilometres
Bonds Quick Check
Metallic bonds are responsible for many properties of metals, such as conductivity. Why is this possible? (1 point)
A Metals have low electronegativity, so they are conductive because they pass charges easily.
Two metals bonded together are going to be more conductive than a metal bonded with a
nonmetal
C An attraction between a positive charge and a negative charge is created.
The bonds can shift because valence electrons are held loosely and move freely.
Answer: The bonds can shift because valence electrons are held loosely and move freely.
Explanation:
the masses of your hand and your notebook are quite small, so the force of attraction between them is
una caja en reposo se traslada 93 cm con un peso de 67N en un tiempo de 9,89h.¿cual es la aceleración la masa y la fuerza de dicho objeto
Answer:
a. Acceleration, a = 1.47 * 10^{-9} m/s²
b. Mass = 4.57 * 10^{10} kilograms
c. Force = 67.12 Newton
Explanation:
Given the following data;
Distance = 93 cm to meters = 93/100 = 0.93 meters
Weight = 67 N
Time = 9.89 hours to seconds = 35604 seconds
Initial velocity = 0 m/s (since it's starting from rest)
Acceleration due to gravity, g = 9.8 m/s
a. To find the acceleration, we would use the second equation of motion;
[tex] S = ut + \frac{1}{2} at^{2} [/tex]
Where;
S is the distance covered or displacement of an object.
u is the initial velocity.
a is the acceleration.
t is the time.
Substituting the values into the equation, we have;
[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]
[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]
[tex] 0.93 = 633822408a [/tex]
[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]
Acceleration, a = 1.47 * 10^{-9} m/s²
b. To find the mass
Weight = mass * acceleration due to gravity
67 = mass * 1.47 * 10^{-9}
[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]
Mass = 4.57 * 10^{10} kilograms
c. To find the force;
Force = mass * acceleration
Force = 4.57 * 10^{10} * 1.47 * 10^{-9}
Force = 67.12 Newton
A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's radius is 30 mm and the tangential velocity is 0.5 m/s?
Answers:
a. 8.33 m/s^2
b. 0.5 m/s^2
c. 30 mm/s^2
d. 0.33 m/s^2
e. 83.20 m/ft^2
Answer:
a
Explanation:
a_c = v_t^2/r
a_c = (0.5)^2/0.03
a_c = 8.33 m/s^2
A lady walks 10 m to the north, then she turns and continues walking 30 m due east.
Determine her(a) distance covered
(b) displacement.
Answer:
The distance covered is 40 m and the displacement is 31,6m.
Explanation:
The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
A force of 5N accelerates a mass by 2 m/s². What will be the acceleration if the force and mass were increased to twice their original value?
Answer:
4m/s4
Explanation:
please answer
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled the last 120 km in 3.2 hours. calculate the average speed of the car during the journey.
average speed of the car is 23.9 km/h
Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east
difference between effort distance and load distance
Answer:
Lever systems are simple machines that change or increase the input force that we apply to the load. The lever provides us with some mechanical...
Answer:
● Effort arm or Effort distance (ED): The perpendicular distance from the fulcrum to the point of effort is called effort arm.
● Load arm or Load distance (LD): The perpendicular distance from the fulcrum to the point of load is called load arm.
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron
Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
State the relative position for the earth and sun in a lunar eclipse (in a partial and total eclipse)
Answer:
A lunar eclipse is when the Earth passes between the moon and the sun, casting a shadow on the moon. This can only occur when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle.
Can someone do this for the football
Estimated volume
(cm3)
Estimated density
(g/cm3)
edge.
Answer:
0.10 g/cm3
TRUST ME GUYS
Có một số điện trở giống nhau R0 = 3
. Cần ít nhất bao nhiêu điện trở để có một
đoạn mạch có điện trở Rtđ = 8
Answer:
hlo
Explanation:
hlo olz mark me as brainlest
write down any 5 example of conservation of momentum?
Answer:
1) Motion of air mass moving from equator northward (closer to earth axis)
2) Motion of object in orbit
3) Collision of 2 objects
4) Skater changing rotation by extension of arms
5) Motion of rocket due to velocity of expelled gas
Một vật không mang điện sẽ bị nhiễm điện dương khí
Answer:
không có điện
Explanation:
An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
transition metals
noble gases
metalloids
alkali metals
Answer:
Alkali metals
Explanation:
Elements in this group are highly reactive, soft, lustrous and highly conductive.
An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
The single valance electron leads to weak bonds amongst the element's atoms which makes them softThe elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.Examples of alkali electrons include:
Lithium Sodium Potassium etcIn conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
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4. When setting goals, you should do everything EXCEPT which of the following?
Take into account your current level of activity.
Ask all of your friends what they think your goals should be..
Set a time limit.
Be realistic.
Question in the picture please help me...
Show all steps please....
Answer: i)A to B : (ice) freezing
ii) B to C (water) boiling
C to D (steam) evaporating
explanation: 0° is the freezing point of water when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates
What is your wheel and axle
Explanation:
The wheel and axle is a type of simple machine used to make tasks easier in terms of manipulating force by applying the concept of mechanical advantage.
Động vật nào sau đây máu đi nuôi cơ thể không pha trộn giữa máu giàu O2 và máu giàu CO2?
A.
Bò sát, chim, thú
B.
Cá, bò sát, chim
C.
Cá, lưỡng cư, bò sát
D.
Cá, chim, thú
Answer:
D
Explanation:
Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn
