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2 years ago

How is force related to math

Physics

2 answers:

dybincka [34]2 years ago

8 0

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

OLga [1]2 years ago

7 0

Answer:

Force is push or pull. Unbalanced forces make an object accelerate. The unit of force is the Newton (abbreviation is N). A Newton is the force it takes to make 1 kg change its velocity by 1 m/s (meter per second) every second.

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If two forces are in the same direction then do they cancel each other out

stellarik [79]

Im pretty sure that they do

3 0

3 years ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram

olya-2409 [2.1K]

Answer:gay

Explanation:

gay

8 0

3 years ago

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

$\theta =1.22\cdot \frac{\lambda }{D}$

Applying the given values we get

$\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads$

thus the linear separation equals $L=\theta \times Distance$

Applying the given values we get

$L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes$

6 0

3 years ago

Introduces una masa de 80g en una probeta con 60cm cúbicos de agua y el nivel sube hasta 75cm cúbicos. qué densidad tiene la mas

ExtremeBDS [4]

Answer:

$d=5.33\ gr/cm^3$

Explanation:

<u>Densidad</u>

La densidad de una masa m que ocupa un volumen v se define como:

$\displaystyle d=\frac{m}{V}$

Tenemos que una masa de 80 gr hace que un volumen de agua se desplace desde 60 cm3 hasta 75 cm3, es decir, el volumen propio de la masa es m = 75 - 60 = 15 cm3.

Con esta información se calcula la densidad:

$\displaystyle d=\frac{80\ gr}{15\ cm^3}$

Operando:

$\boxed{d=5.33\ gr/cm^3}$

4 0

3 years ago