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3 years ago

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The formula for force (CBSE 8TH GRADE - Physics)

1 answer:

Tanzania [10]3 years ago

8 0

Answer:

F=ma

Explanation:

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Thought Experiment: A monkey escapes from a zoo and climbs a tree. After failing to entice the monkey down, a zookeeper fires a

Andreyy89

Explanation:

When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"

so we can find the time to reach the bullet to the monkey

$t = \frac{d}{vcos\theta}$

Now similarly we can find the vertical displacement of the bullet in the same time

$\Delta y = vsin\theta t - \frac{1}{2}gt^2$

$\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2$

so it is given as

$\Delta y = d tan\theta - \frac{1}{2}gt^2$

here if the monkey is initially at height H above the ground at given angle then we can say

$H = dtan\theta$

so we can say that

$\Delta y = H - \frac{1}{2}gt^2$

So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as

$\Delta y = H - \frac{1}{2}gt^2$

so here bullet will hit the monkey as both monkey and bullet are at same position.

3 0

3 years ago

A particle travels along a straight line with a velocity v = (12 - 3t) m/s, where t is in seconds. When t = 1 s, the particle is

Simora [160]

Answer:

(1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

Explanation:

Given that,

Velocity = (12-3t^2) m/s

When t = 1 s, the particle is located 10 m to the left of the origin.

We need to calculate the acceleration at t = 4 sec

Using formula of acceleration

$a=\dfrac{dv}{dt}$

Put the value of v

$a=\dfrac{d}{dt}(12-3t^2)$

$a=-6t$

Put the value of t

$a=-6\times4$

$a=-24\ m/s^2$

The displacement from t = 0,t = 10 s, and the distance the particle travels during this time period.

We need to calculate the distance

Using formula of distance

$ds=v\ dt$

$\int_{-10}^{s}=\int_{1}^{t}{v}dt$

Put the value of v

$\int_{-10}^{s}=\int_{1}^{t}{ (12-3t^2)}dt$

$s+10=12t-t^3-11$

$s=12t-t^3-21$

At t = 0,

$s_{t=0}=-21$

At t = 10,

$s_{t=10}=12\times10-10^3-21$

$s_{t=10}=-901$

The displacement is

$\Delta s=-901-(-21)$

$\Delta s=-880\ m$

The distance at t= 2 sec

$s_{t=2}=12\times2-2^3-21$

$s_{t=2}=-5$

The total distance will be,

$s_{T}=(21-5)+(901-5)$

$s_{T}=912\ m$

(2). We need to calculate the distance at 2 sec

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$s=27\times 2+\dfrac{1}{2}\times6\times(2)^3$

$s=78\ m$

Hence, (1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

3 0

3 years ago

How is the duration and intensity of daylight related to the suns apparent path?

DedPeter [7]

Answer:

I dont know but ask a tutor

Explanation:

MOnkey Srry

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2 years ago

Hannah wants to write a book about how scientists and society interact, and she has generated ideas for chapters. Which chapter

tigry1 [53]

Number three. How water evaporates and then forms rain and snow.

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3 years ago

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The amount of work done depends on what two things

fomenos

mass and acceleration have a nice day

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3 years ago