Ask question

Ask question

All categories

3 years ago

15

A dancer lifts his partner above his head with an acceleration of 2.5 m/s squared the dancer. Exerts a force of 200N what is the

mass of the partner

2 answers:

elena-s [515]3 years ago

4 0

F = ma
F = 200N, a = 2.5m/s^2

200N = (2.5)m
m = 80kg

shepuryov [24]3 years ago

3 0

Answer:

mass = 80 kg

Explanation:

As per Newton's II law we know that

$F = ma$

here we know that

$F = 200 N$

also we know that

$a = 2.5 m/s^2$

so from above formula we have

$200 = m(2.5)$

$m = \frac{200}{2.5}$

$m = 80 kg$

You might be interested in

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

a net force of 100 newton’s is a moving mass with an acceleration of 5 m/s. what is the mass of the object?

Jlenok [28]

F = ma

100 N= m . 5 ms⁻¹

100 / 5 = m

20 N ms = mass of object.

7 0

3 years ago

Calculate the vapor pressure for a mist of spherical water droplets of radius 3.70×10−8m surrounded by water vapor at 298 K. The

poizon [28]

Answer:

The vapor pressure for a mist is $P= 25.92\ Torr$

Explanation:

From the question we are given that

The radius is $r = 3.70 *10^{-8} m$

The temperature is $T = 298K$

The vapor pressure of water $P_o = 25.2\ Torr$

The density of water is $\rho = 998 kg.m^{-3}$

The surface tension of water is $\sigma = 71.99 m N \cdot m^{-1}$

Generally the equation of that is mathematically represented as

$ln (\frac{P}{P_0} ) = \frac{2 \sigma M}{r \rho RT}$

Where P is the vapor pressure for mist

R is the ideal gas constant = 8.31

making P the subject in the formula

$P = e^ {\frac{2 \sigma M}{r \rho RT}} * P_0$

$= e^{\frac{2 *(0.07199)(0.018)}{(3.70*10^{-8})(998)(8.31)(298)} } * 25.2$

$P= 25.92Torr$

4 0

3 years ago

How was the Periodic Table of Elements developed and how are the elements arranged on it?

victus00 [196]

Answer:

In 1869 Russian chemist Dimitri Mendeleev started the development of the periodic table, arranging chemical elements by atomic mass. He predicted the discovery of other elements, and left spaces open in his periodic table for them.

Explanation:

4 0

3 years ago

(I) A car slows down from 28 m????s to rest in a distance of 88 m. What was its acceleration, assumed constant?

Nataly [62]

Answer:

The value is $a = - 4.45 m/s^2$

Explanation:

From the question we are told that

The initial speed is $u = 28 \ m/s$ at a distance of $s_1 = 0 \ m$

The final speed is $v = 0 \ m/s$ at a distance of $s_2 = 88 \ m$

Generally from the kinematic equation we have that

$v^2 = u^2 +2as$

=> $a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}$

=> $a = \frac{0 - 28^2 }{ 2(88 - 0 )}$

=> $a = - 4.45 m/s^2$

The negative sign shows that it is decelerating

6 0

4 years ago