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3 years ago

If the net force on a block is zero

Physics

1 answer:

amm18123 years ago

8 0

If the net force on a block is zero, the block will move at constant velocity

Explanation:

We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

$\sum F = ma$ (1)

where

$\sum F$ is the net force on the object

m is its mass

a is its acceleration

In this problem, we have a block, and the net force on it is zero:

$\sum F = 0$

According to eq.(1), this also implies that

$a=0$

So, the acceleration of the block is zero.

However, acceleration is the rate of change of velocity of a body:

$a=\frac{\Delta v}{\Delta t}$

where $\Delta v$ is the change in velocity in a time of $\Delta t$ . Since the acceleration is zero, this means that $\Delta v=0$ , and therefore the velocity of the object is constant.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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Consider the following three concentric systems two thick shells and a solid sphere all conductors The radii in the increasing o

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Answer:

Explanation:

From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.

So, in light of that, to determine the electric charges values & signs on each of them, we have:

on a = +3 C

on b = -7 C

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on d = +3 C

on e = -7 C

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3 years ago

Which example involves the transformation of chemical energy directly into light energy?

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burning sodium or magnesium

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3 years ago

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

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3 years ago

Read 2 more answers

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0

3 years ago

A ball rolls off a desk at a speed of 2 m/s and lands 0.50 seconds later.

Hoochie [10]

Answer:

(a) The ball lands <u>1 m</u> ahead of the desk's base.

(b) The height of the desk is <u>1.225 m.</u>

Explanation:

Given:

Initial velocity of the ball is, $u_{0}=2\ m/s$

Time of flight of the ball is, $t=0.50\ s$

(a)

Let the ball fall at a distance of $x$ from the base of the desk and let the height of the desk be $y$ .

The given motion of the ball is a projectile motion that can be divided into 2 mutually perpendicular directions.

Now, considering only the horizontal motion of the ball. The ball not under the influence of any force in the horizontal direction.

Therefore, the distance covered by the ball horizontally is given as:

$x=v_0\times t\\x=2\times 0.50=1\ m$

So, the ball lands 1 m ahead of the desk's base.

(b)

Now, consider only the vertical motion of the ball. The ball is under the influence of gravity.

So, vertical distance can be obtained using Newton's equations of motion.

We use the following equation of motion:

$y-y_0=u_{oy}t+\frac{1}{2}gt^2$

Where,

$y_{0}\rightarrow \textrm{initial vertical position of ball}\\u_{oy}\rightarrow \textrm{initial vertical velocity}\\g\rightarrow \textrm{acceleration due to gravity}\\y\rightarrow \textrm{final position of ball vertically}$

Here, as the ball leaves the desk horizontally, initial velocity is only in the horizontal direction and doesn't have any component in the vertical direction. So, $u_{oy}=0\ m/s$

Plug in $y_0=h,y=0,g=-9.8,u_{oy}=0,t=0.5$ . Solve for $h$ .

$0-h=0+\frac{1}{2}(-9.8)(0.5)^2\\-h=-4.9\times 0.25\\h=1.225\ m$

Therefore, the height of the desk is 1.225 m.

6 0

3 years ago