a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in the trampoline

Answer:

The theory of relativity usually encompasses two interrelated theories by Albert Einstein: special relativity and general relativity, proposed and published in 1905 and 1915, respectively. Special relativity applies to all physical phenomena in the absence of gravity. General relativity explains the law of gravitation and its relation to other forces of nature.It applies to the cosmological and astrophysical realm, including astronomy.

The theory transformed theoretical physics and astronomy during the 20th century, superseding a 200-year-old theory of mechanics created primarily by Isaac Newton. It introduced concepts including spacetime as a unified entity of space and time, relativity of simultaneity, kinematic and gravitational time dilation, and length contraction. In the field of physics, relativity improved the science of elementary particles and their fundamental interactions, along with ushering in the nuclear age. With relativity, cosmology and astrophysics predicted extraordinary astronomical phenomena such as neutron stars, black holes, and gravitational waves

Convert to m/s

[tex]\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s[/tex]

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

[tex]\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=a=-10m/s^2[/tex]

Using second equation of kinematics

[tex]\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2[/tex]

[tex]\\ \sf \longmapsto s=40+(-20)[/tex]

[tex]\\ \sf \longmapsto s=40-20[/tex]

[tex]\\ \sf \longmapsto s=20m[/tex]

Now

Using newtons second law

[tex]\\ \sf \longmapsto Force=ma[/tex]

[tex]\\ \sf \longmapsto Force=5000(-10)[/tex]

[tex]\\ \sf \longmapsto Force=-50000N[/tex]

[tex]\\ \sf \longmapsto Force=50kN[/tex]

Answer . The acceleration of the truck is 10m/[tex]s^{2}[/tex], and the distance covered is 40 m. Have attached the picture for solution.

Hope that helps.

12 x cos 50 = 7.713451316...

Answer:

This type of a behavior is known as oscillation, a periodic movement between two points. ... Amplitude: The distance from the center of motion to either extreme. Period: The amount of time it takes for one complete cycle of motion.

Explanation:

Amplitude (a):- The maximum displacement of particle from its mean position on either side is called amplitude.  

Periodic time:- The time taken by a wave to complete one oscillation is called periodic time.

The answer is definitely Potassium

Explanation:

speeds = distance/time

=100/16

=6.25m/s

second speed is;

400/44 =9.09

av. speed = total speed /n

= (6.25+9.09)/2

=7.67 m/s

Answer:

increase in mass of body

Answer:

X = 24.135 kilometres

Explanation:

Given the following data;

Distance = 15 miles

To convert the value in miles to kilometers;

Conversion:

1 mile = 1.609 kilometres

15 miles = X kilometres

Cross-multiplying, we have;

X = 1.609 * 15

X = 24.135 kilometres

it takes 6 hours to travel back to Hyderabad

Answer:

The momentum makes it worse.

Explanation:

The momentum of vehicles running at faster speeds is very high and causes a lot of damage to the vehicles.

1:-

[tex]\\ \sf\longmapsto cos\Theta=\dfrac{B}{H}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\dfrac{4}{5}[/tex]

2:-

[tex]\\ \sf\longmapsto tan\Theta=\dfrac{P}{B}[/tex]

[tex]\\ \sf\longmapsto Tan\Theta=\dfrac{3}{4}[/tex]

Key Notes:-

Answer:

Cos A = 4/5

Tan A = 3/4

Explanation:

Cos A = Adjacent/Hypotenuse = 4/5

Tan A = Opposite/Adjacent = 3/4

Answer:

Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers

Mathematically, we have;

[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]

Where;

m₁, and m₂ are the masses of the bodies

R = The distance between their centers

G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²

The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers

Q. 2. Newton's law of gravitation in vector form is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]

The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1

The force of gravity of body 2 on 1 is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]

The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)

[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]

Explanation:

Answer:

Efficiency of a machine is defined as the ratio of output work to input work in a machine . It is expressed in percentage and denoted by

η ( eta).

Answer:

1) Can change the state of an object(rest to motion/ motion to rest):For example, pushing a heavy stone in order to move it. 2) May change the speed of an object if it is already moving. 4) May bring about a change in the shape of an object. For example, blowing air in balloon.

Explanation:

Effect of force Can change the state of an object(rest to motion/ motion to rest):For example, pushing a heavy stone in order to move it.

Answer:

well done buddy

Explanation:

Answer:

The SI unit of upthrust is Newton(N).

The SI unit of preesure is Pascal(P).

Thank You

Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements

(a) The maximum error in D is 0.09

(b) The standard error in D is approximately 0.034

The procedure for arriving at the above values is as follows;

The given measurements and the sited errors are;

A = 15.21 + 0.01

B = 10.82 + 0.05

C = 11.00 + 0.03

D = A + B + C

(a) Required parameter;

To calculate the maximum error in D

The equation for the propagation of error in addition is presented as follows;

Given that we have;

x = a + b

Therefore;

x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)

∴ Δx = Δa + Δb

From the above formula, we have;

Where;

D = A + B + C

The maximum error in D = The sum of the maximum error in A, B, C

∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09

(b) Required parameter:

To find the standard error in D

The standard error is the sampling distribution's standard deviation, SD

Variance = SD²

The combined variance, SD² = The sum of the squares of individual standard deviations

Given that the standard errors represents the standard deviation, we get;

The combined variance, SD² = 0.01² + 0.05² + 0.03²

The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059

[tex]Standard \ error = \dfrac{SD}{\sqrt{n} }[/tex]

Where n = 3, for the three measurement, we get;

[tex]Standard \ error = \dfrac{\sqrt{0.01^2 + 0.05^2 + 0.03^2} }{\sqrt{3} } \approx 0.034[/tex]

The standard error in D is approximately 0.034

Learn more about maximum error and standard error here:

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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

Answer:

A concave mirror, or converging mirror, has a reflecting surface that is recessed inward (away from the incident light). Concave mirrors reflect light inward to one focal point. They are used to focus light.

Answer:

Zero because the applied force is perpendicular to the motion of the object.

No work is done on an object moving is a circular path about a central attractive force.

Any work done in such a case would result in a change in the orbit.

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

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Conversion:

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

Answer:

the gas that escaped from the mixture contained the missing weight

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

please click thanks and mark brainliest if you like :)

Answer:

a) cos30=adj/hyp

cos30= horizontal force/10

horizontal force= 8.66 N

rest of a is in the picture.

b) i believe you can continue.

We have a problem about conservation and velocity, we will find that it does affect the speed of the ball, increasing it by 1.7m/s.

There is something called momentum, which we can define as the "quantity of movement", and we can simply write as the product between velocity and mass.

The momentum is conservative, then we have conservation of momentum.

This means that when you run whit the ball in your hands, the momentum of the ball will be equal to your velocity times the mass of the ball, and this must conserve after you throw the ball.

Now with this idea in mind, this means that if you run with a velocity V, and you throw the ball with a velocity V', the velocity of the ball when it leaves your hand will be:

V + V'.

So, if you run with a velocity of 1.7m/s forward and you throw the ball (assuming in the same direction) the speed of the ball will be 1.7m/s larger than if you were to throw it standing still.

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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]

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