Which describes a force acting on an object?

You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist

omeli [17]

The acceleration of the ball after leaving the hand is $9.8 m/s^2$ downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

$F=mg$

where m is the mass of the ball and g is the acceleration of gravity ( $g=9.8 m/s^2$ ), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

$a=\frac{\sum F}{m}$

where

$\sum F$ is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

$a=\frac{mg}{m}=g=9.8 m/s^2$

This means that the acceleration of the ball remains $9.8 m/s^2$ downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

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5 0

3 years ago

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

so

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

3 years ago

The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.

34kurt

You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be
1 km = 0.6214 miles 
1000 m = 1 km 
60 seconds = 1 minute 
60 minutes = 1 hour. 
2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?

6 0

4 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago

What's the value of 57,281 joules in Btu? A. 54.3 Btu B. 14.2 Btu C. 28.9 Btu D. 37.7 Btu

Alex Ar [27]

I'm quite sure it is A

8 0

4 years ago

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