find the area of a right triangle with sides 15.0 cm, 20.0 cm, and 25.0 cm. express your answer with the correct number of significant figures

Answers

Answer 1

Answer: The area of this triangle can be calculated using herons formula since th three sides are given. It is expressed as:

A=sqrt( s(s-a)(s-b)(s-c))

where s is equal to a+b+c / 2

s=15 +20 +25 /2=30
A=sqrt( 30(30-15)(30-20)(30-25))
A= 150 cm^3

Related Questions

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????
Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released
An earthquake on the ocean floor produced a giant wave called a tsunami. The tsunami traveled through the ocean and hit a remote island, causing a lot of damage. Is the water that hit the island the same water that was above the earthquake on the ocean floor?A No, the water from above the earthquake stayed in the same place and only the energy was transferred.B No, the energy in the wave pushed the water particles from above the earthquake in the opposite direction.C Yes, the water particles moved toward the island while the energy remained above the earthquake.

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of the object's acceleration by what factor?

Answers

Answer:

acceleration will be tripled.

Explanation:

We know, when an object is performing Simple harmonic motion, the force

experience by it is directly proportional to its displacement from its mean position.

Also, F = ma , therefore, acceleration is also proportional to its displacement .

Now, F = kx

Therefore, $a=(k\ x)/(m)$

If we triple the displacement i.e, 3x.

Acceleration $a'=(k(3x))/(m)=3a.$

Therefore, acceleration is also tripled.

Hence, this is the required solution.

You are given a parallel plate capacitor with plates having a rectangular area of 16.4 cm2 and a separation of 2.2 mm. The space between the plates is filled with a material having a dielectric constant κ = 2.0.Find the capacitance of this system

Answers

Answer: C = 1.319×10^-11 F

Explanation: The formulae that relates the capacitance of a capacitor to the area of the plates, distance between the plates and dielectric constant is given as

C = kε0A/d

Where C = capacitance of plates =?

k = dielectric constant = 2.0

Area of plates = 16.4cm² = 0.00164 m²

d = distance between plates = 2.2 mm = 0.0022m

By substituting the parameters, we have that

C = 2 × 8.85×10 ^-12 ×0.00164/ 0.0022

C = 0.029028 × 10^-12/ 0.0022

C = 13.19× 10^-12

C = 1.319×10^-11 F

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

Answers

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^(-8) J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^(-8))(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km west

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

North component = 835 km/h * sin(41.5°)
West component = 835 km/h * cos(41.5°)
Distance north = North component * 2.20 h
Distance west = West component * 2.20 h

Learn more about Velocity components here:

brainly.com/question/14478315

#SPJ3

Other Questions

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.When the rock is released, what is its kinetic energy?

If the_____of a wave increases, its frequency must decrease.a. period b. energy c. amplitude d. velocity