Answer:Yes, A body can have constant speed but still accelerate as in case of uniform circular motion. In uniform circular motion speed remains constant and direction of velocity changes with every point in the direction of tangent drawn from that point.
Explaination:
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a small object is placed between two plane mirrors inclined at an angle of 60° to each other in a dark room how many images are seen explain
Answer:
nothing
Explanation:
bocouse of darkness
7. You are using a Bunsen burner to heat a chemical. You need your notebook, which is on the other side of the flame.
Accident:
Prevention:
Accident: Get burned
Prevention: turn off the burner.
sort out electric current as fundamental or derived unit.
Answer:
electric current is derived unit.
Explanation:
According to the definition of electric current, it appears to be a derived quantity. Charge on the other hand seems more fundamental than electric current.
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
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
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
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
What unit is used in MKS system and FPS system
The "second" is the base unit of time in both systems.
What is the minimum value of force acting between two charges placed at 1 m apart from each other?
(a)Ke²
(b)Ke
(c)Ke/4
(d)Ke² /2
Answer:
Ke²
Explanation:
So,
q1 = e
q2 = e
r = 1m
By coulumb's law,
F = K (q1q2/r²)
F = K (e)(e)/(1)²
F = Ke²
Option(a)
a car takes 10 minutes to travel 10 km calculate average speed of car.
Answer:
(km/mins) × ( mins/hr) = km/hr
(10/10)×(60/1) =600/10 = 60 km /hr
......
A ball is launched from the ground with the horizontal speed of 30 m/s and vertical speed of 30 m/s what angle was the ball launch at?
Answer:
45 degrees
Explanation:
angle of launch=arctan(vertical velocity/ horizontal velocity)
angle = arctan(30/30) = 45 degrees
s27253129 ,,, message me please, I can't ask you my homework question in the comments :c
Define Metrology
define Metrology
Answer:
the scientific study of measurement.
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
Numerical problems:
a. convert the following as instructed:
i) 340 cm into m
ii)86400 seconds into day
Answer:
a=3.4m because of the m
b=1day because 86400=a day
giving me the points are enough
Answer:
the product of mass and velocity
....in my syllabus
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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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.
3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided there is no loss of heat in the surrounding. the specific heat capacity is 4200j/kg
Answer:
hope fully it help s
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^2Que. 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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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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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
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
How would the period of this pendulum differ from an equivalent one on earth?
Answer:
the pendulum differs from 300 inches
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
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
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.
what are the examples of Inertia of motion
Seat belts tighten in a car when it stops quickly.
Men in space find it more difficult to stop moving because of a lack of gravity acting against them.
When playing football, a player is tackled, and his head hits the ground.
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