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2 years ago

Define the term acceleration due to gravity

Physics

1 answer:

kifflom [539]2 years ago

7 0

Answer

Acceleration

Acceleration Means rate of Change of velocity with respect to time.

acceleration due to gravity

when a body falls freely means no other force act on it except gravitational force that time velocity of object does not remains constant but it changes as it approaches towards the ground it means body get accelerated and this acceleration which is gained by the body due to gravitational force is called its acceleration due to gravity it is denoted by g. Acceleration due to gravity is different on different planet. SI unit of is m/s^2 . On earth its standard value is 9.80665m/s^2.

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In the diagram, ∠D is b

Degger [83]

Answer:

The answer is B

Explanation:

3 0

3 years ago

A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and

kvv77 [185]

Answer:

(a) 1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

$\overrightarrow{d_{1}}=580\widehat{j}$

$\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}$

$\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}$

The resultant displacement is given by

$\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}$

$\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}$

magnitude of the displacement

$d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m$

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

$tan\theta =\frac{1294.18}{35.36}=36.6$

θ = 88.44°

4 0

3 years ago

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo

Dafna1 [17]

Answer: 75 N to the right

Explanation:

5 0

2 years ago

Which physical activity would a global positioning system resource help you with?

PolarNik [594]

The physical activity that a global positioning system resource can help me with is : ( A ) Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is Hiking.

Learn more about GPS : brainly.com/question/9795929

#SPJ1

4 0

1 year ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago