The speed of the cat

What is the mass of a man who accelerates 4 m/s2 under the action of a 200 N net force?

Over [174]

Answer:

$\huge \boxed{ \boxed{50 \: kg }}$

Explanation:

The mass of the man can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{200}{4} \\$

We have the final answer as

Hope this helps you

7 0

2 years ago

Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick

77julia77 [94]

Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)

Solids' molecules are strongly attracted to one another. As a result, the molecules are barely moving and tightly packed. Because of this, shape and volume are fixed.

The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).

You can learn more about states of the matter here:

brainly.com/question/18538345

#SPJ4

6 0

2 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$