when a body moves from point A to point B the distance travelled in 500 meters with in a given time of 0.6 hours and find out the velocity and acceleration in a body.


what is correct answer .​

Answers

Answer 1

Answer

Explanation:

Convert the time to seconds = 0.6 × 60 × 60

= 2160seconds

Velocity = distance ÷ time

Velocity = 500 ÷ 2160

Velocity = 0.23meters per seconds(m/s)

Acceleration = Velocity ÷ time

Acceleration = 0.23 ÷ 2160

Acceleration = 0.000106meters per seconds ²(m/s²)


Related Questions

The mass of objects is 4kg and it has a density of 5gcm^-3. what is the volume ​

Answers

Answer:

4kg×5gm^3=60

Explanation:

the object if heavy

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

Answers

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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What is the connection of H ions at a ph=2?

Answers

Answer:

Explanation:

High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). The overall concentration of hydrogen ions is inversely related to its pH and can be measured on the pH scale

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.

Answers

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) .

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 ?

Answers

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.

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 ​

Answers

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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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?

Answers

Answer:

4m/s4

Explanation:

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.

Answers

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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12 x cos 50 = ?

Does anyone have the answer ? I forgot my my calculator.

Answers

12 x cos 50 = 7.713451316...

the answer is 7.713451316

Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m​

Answers

Time=1s=tAcceleration due to gravity=g=9.8m/s^2Distance=s

In free fall

[tex]\boxed{\sf s=-\dfrac{1}{2}gt^2}[/tex]

[tex]\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2[/tex]

[tex]\\ \sf\longmapsto s=-4.9(1)[/tex]

[tex]\\ \sf\longmapsto s=-4.9m[/tex]

Take it positive

[tex]\\ \sf\longmapsto s=4.9m[/tex]

Option a is correct

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.)

Answers

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

giving me the points are enough

Answers

Answer:

the product of mass and velocity

....in my syllabus

an object is sliding down in clean plane the velocity change at a constant rate from 10 cm to 15 CM in 2 second what is it acceleration ?​

Answers

Initial velocity=10m/s=u

Final velocity=v=15m/s

Time=t=2s

[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{15-10}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{5}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=2.5m/s^2[/tex]

Answer: a = 2.5 cm/s²

Explanation:

Acceleration = (Final velocity - Initial Velocity)/time taken

a = v-u/t

Initial velocity = 10 cm/s

Final velocity = 15 cm/s

Time = 2 seconds

a = (15-10)/2

a = 5/2

a = 2.5 cm/s²

Therefore the acceleration is 2.5 cm/s²

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determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3​

Answers

Answer:

divide the density of solution by density of water

EXPLANATION:

LIKE:

1.25÷1000kgm-3

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

Answers

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

Draw a wave that has a wavelength of 3 cm and an amplitude of 1 cm. Label the wavelength, the amplitude, the rest position, and the crest and trough of your wave.

Answers

Answer:

Please find attached, the required wave drawn with MS Excel

Explanation:

Functions that represent waves is given as follows

A general form of the wave equation is A·sin(B·x) + D

Where;

B = 2·π/T

T = The period of the wave = 1/f

D = The vertical shift of the wave = 0

A = The amplitude of the wave = 1 for sine wave

v = The wave velocity

λ = The wavelength of the wave

f = The frequency of the wave

v = f·λ

At constant v, λ ∝ 1/f  

∴ λ ∝ T

Where T = 3, we have;

B = 2·π/T

∴ B = 2·π/3

Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows

y = sin((2·π/3)·x)

Plotting the above wave with MS Excel, we can get the attached wave

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.

Answers

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.

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.

Answers

Answer: The bonds can shift because valence electrons are held loosely and move freely.

Explanation:

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?

Answers

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

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

Answers

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

Answers

Reference frame definitely changes when the body is changing. That is the reason that in order to describe the position of a point that moves relative to a body that is moving relative to the Earth, it is usually convenient to use a reference frame attached to the moving body.

the masses of your hand and your notebook are quite small, so the force of attraction between them is

Answers

The force of attraction is decreased

Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed

Answers

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

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​

Answers

Explanation:

based on the above information

1.A

2.B

3. C

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

Answers

The answer is definitely Potassium

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.

Answers

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^2

5. a. Answer the following questions. What is density? Write a formula by showing the relation among density mass and volume.​

Answers

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.

round off 20.96 to 3 significant figures. a.20.9 b.20 c.21.0 d.21​

Answers

Answer:

option c. 21.0

Explanation:

It was given that to find 3 significant figures. So the answer is 21.0

sl unit of upthrust and SI unit of pressure​

Answers

Answer:

The SI unit of upthrust is Newton(N).

The SI unit of preesure is Pascal(P).

Thank You

What unit is used in MKS system and FPS system​

Answers

The "second" is the base unit of time in both systems.

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