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3 years ago

What is the formula for calculating acceleration

Physics

1 answer:

Komok [63]3 years ago

7 0

Answer:

the formula for calculating acceleration is ending speed minus starting speed divided by time.

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When you are sitting in a chair, your body exerts a _____

Vikki [24]

When you are sitting in a chair your body exerts a FORCE, the chair exerts the SAME force back.

4 0

3 years ago

Please help! The image produced by a concave mirror is ? .

Alexeev081 [22]

Answer:

is a reflection.

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.

Explanation:

6 0

3 years ago

Read 2 more answers

What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is

ss7ja [257]

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R² ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

3 0

3 years ago

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0

4 years ago

At the end of a delivery ramp, a skid pad exerts a constant force on a package so that the package comes to rest in a distance d

MatroZZZ [7]

Answer:

increases by a factor of $\sqrt{2}$

Explanation:

First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the package when it stops, $v_0$ is the initial velocity of the package when it, a is the deceleration, and $\Delta s = d$ is the distance traveled.