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jasenka [17]
3 years ago
7

A roller coaster accelerates from 0 m/s to 25 m/s in 5 seconds. Determine the acceleration of the roller coaster.

Physics
1 answer:
pashok25 [27]3 years ago
5 0
Acceleration = change of velocity / time taken = 25/5 = 5 m/s/s
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which of the following is the correct si unit to use in measuring the mass of a boulder a. tons b. kilograms c. ounces d. pounds
Salsk061 [2.6K]
B. kilograms iuyiuiuhikj
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3 years ago
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Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the c
Darina [25.2K]

Answer:

bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon

Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

5 0
2 years ago
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba
Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

\mu = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

m_1 = Mass of bean bag = 0.354 kg

m_2 = Mass of empty crate = 3.77 kg

v_1 = Speed of the bean bag

v_2 = Speed of the crate

Acceleration

a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g

a=--9.81\times 0.48=4.7088\ m/s^2

From equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s

In this system the momentum is conserved

m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s

The speed of the bean bag is 31.42383 m/s

8 0
3 years ago
Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi
Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0
3 years ago
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25
Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: d_{1x} = 5.2 cos 25^{\circ}=4.7 mi

North: d_{1y} = 5.2 sin 25^{\circ}=2.2 mi

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: d_{2x}=0

North: d_{2y} = 5.0 mi

So the components of the total displacement are

East: d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi

North: d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi

5 0
3 years ago
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