Answer:600N/m
Explanation: Given ;
Applied force(F) = 180N
Total extension (l) = 30cm =0.30m
Spring constant(K) =?
From Hooke's law
F= kx
therefore k= F/x
k = 180N/0.3m
600N/m
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
La resistencia de un termómetro de platino es de 6Ω a30°C. Hallar su valor correspondiente a 100°C,sabiendo que el coeficiente de temperatura de resistividad del platino vale 0,00392°C^(-1).
Respuesta:
7,6 Ω
Explicación:
Paso 1: Información dada
Resistencia a 30 °C (R₀): 6 ΩCoeficiente de temperatura (α): 0,00392 °C⁻¹Paso 2: Hallar la resistencia (R) a 100 °C
Podemos ver la relación entre la resistencia de un material y la temperatura usando la siguiente ecuación.
R = R₀ (1 + α × ΔT)
R = 6 Ω (1 + 0,00392 °C⁻¹ × (100 °C - 30 °C)) = 7,6 Ω
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
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.
The mass of objects is 4kg and it has a density of 5gcm^-3. what is the volume
Answer:
4kg×5gm^3=60
Explanation:
the object if heavy
12 x cos 50 = ?
Does anyone have the answer ? I forgot my my calculator.
12 x cos 50 = 7.713451316...
the masses of your hand and your notebook are quite small, so the force of attraction between them is
A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calculate its final temperature.
We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat. Calculate
the specific heat of the metal.
Answer:
A. DT is given by Q= MCs DT
m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk ×DTk
=250 × 9 × 5
129
=Dt = 180.1085271
answer is 180degree C.
Explanation:
B. = 25×10 ×100
1.082
=2500
1.082
= 23105.360 g/kj.
The final temperature is 180 degree. and the specific heat of the metal is 23105.360 g/kj.
How to calculate the specific heat?Q = m . C . ΔT
Q = heat; m = mass; C is the specific heat and
ΔT = Final T° - Initial T°
Q = C lat . m
Q = Heat
m = mass
C lar = Latent heat of fusion
A) DT is given by Q= M Cs DT
where, m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk × DTk
=250 × 9 × 5
129 =Dt = 180.1085271
Thus, the final temperature is 180 degree.
B) We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat = 25×10 ×100
=2500
1.082
Q = 23105.360 g/kj
Hence, the specific heat of the metal is 23105.360 g/kj.
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Define Metrology
define Metrology
Answer:
the scientific study of measurement.
I 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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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 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) .
giving me the points are enough
Answer:
the product of mass and velocity
....in my syllabus
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
Reference frame definitely changes when also changes
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
What unit is used in MKS system and FPS system
The "second" is the base unit of time in both systems.
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
what ia measurement in science?
= The process of comparing an unknown quantities with an standard known quantities is called measurement.
Yes it is the measurement in science
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^2sl unit of upthrust and SI unit of pressure
Answer:
The SI unit of upthrust is Newton(N).
The SI unit of preesure is Pascal(P).
Thank You
Using your Periodic Table, which of the elements below is most likely to be a solid at room temperature?
A.) potassium, B.) Hydrogen, C.) Neon, D.) Chlorine
The answer is definitely Potassium
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
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:
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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5. a. Answer the following questions. What is density? Write a formula by showing the relation among density mass and volume.
Answer:
Density is how compact something is. The relationship is M/V=D (Mass divided by Volume equals Density).
Explanation:
WHAT IS DENSITY:
Density is the degree of compactness of a substance.
EXAMPLE:
"a reduction in bone density"
FORMULA OF DENSITY:
The formula for density is d = M/V, where d is density, M is mass, and V is volume.
Answer the following questions. 3 A student runs 2 m/s. What does this mean?
Answer:
2ms-¹ means that the body under consideration moves 2m in a second, and may be it will continue to move 2m in every 1 second, if there's no external unbalanced force acting on that body (those forces do include frictional forces). mark its brainlist plz. Kaneppeleqw and 6 more users found this answer helpful. Thanks 3.
Answer:
that the student has travels 2 meters every 1 second that passes
Many people believe that if the human race continues to use energy as we are now, without change, we'll witness a significant worldwide environmental impact in this century. Research this topic and discuss this possibility. Include concrete examples of specific environmental consequences of global warming.
Answer:
It is correct to say that if the human race continues to use energy as it is now, without change, we will witness negative environmental impacts around the world in this century.
As a concrete example, we can cite the means of transport that use fossil fuels, such as cars and buses, which release polluting gases into the atmospheric layer and cause the greenhouse effect, contributing to global warming.
To solve these problems, it is necessary to raise the awareness of individuals, so that there is more and more interest and search for environmentally responsible solutions, such as the large-scale production of electric cars, which do not pollute the environment.
