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3 years ago

you push the ball with a force of 22.8N which induces a -2.3 frictional force what is the net force while you push the ball

1 answer:

4 0

Force is a vector quantity in which direction matters. This means that the forces will act in opposing directions so the net force will be
22.8 - 2.3 =
20.5 Newtons

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

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

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

Help with velocity math pls help asap

pychu [463]

2 is the answer have a nice day <3

5 0

2 years ago

An iron ball weighs 80 N on the surface

Degger [83]

Answer:

2√5N will be the weight.

Explanation:

7 0

3 years ago

You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j

Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

$E=\frac{1}{2} (150) (0.3)^2$

$E=6.75J$

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

#SPJ4

6 0

1 year ago

Elliptical galaxies are frequently found a) Inside the Milky Way b) In galaxy clusters c) In the Galactic bulge d) In the Local

sergejj [24]

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

5 0

3 years ago