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antoniya [11.8K]
3 years ago
8

A jogger moves from a position x =

Physics
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

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A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?
Korvikt [17]

\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

  • u = 250 m/s
  • a = -3 m/s²
  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

4 0
3 years ago
which type of graph shows data seperated into intervals? A. stemplot B. Line Graph C. Scatterplot D. Histogram
USPshnik [31]

Answer:

D. Histogram.

Explanation:

A histogram with equal intervals is suitable here.

8 0
3 years ago
At a particular instant, a proton at the origin has velocity &lt; 5e4, -2e4, 0&gt; m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

9.7\times 10^{-5} T

Explanation:

Velocity =5\times 10^4i-2\times 10^4j

r=0.03i+0.05j

r=\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058

v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

Mass of proton=1.67\times 10^{-27} kg

Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

-The roller doesn't slip while rolling;

v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 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
2 years ago
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