How much work is done if you push a box 200 meters with a force of 35 newtons answer?

Two students play tug of war. Jack pulls East with a force of 82 Newtons while Tony pulls west with a force of 96 Newtons. Which

lorasvet [3.4K]

The net force on the center of the rope is C. 14 N West

Explanation:

We have two forces acting on the rope:

- The force applied by Jack, 82 N, towards East

- The force applied by Tony, 96 N, towards West

Taking East as positive direction, we can rewrite the two forces as

$F_1 = +82 N\\F_2 = -96 N$

where the second force is negative since its direction is west.

The two forces act along the same line, so we can calculate their resultant by simply calculating their algebraic sum. Therefore:

$F=F_1+F_2=+82+(-96)=-14 N$

And the negative sign tells us that the direction is West.

Learn more about forces:

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5 0

4 years ago

Read 2 more answers

What happens when light enters a pair of glasses

alexdok [17]

Answer:

It refracts when it hits the glass.

6 0

3 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

nataly862011 [7]

Answer with Explanation:

Let rest mass $m_0$ at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force= $\frac{GmM}{r^2}$

Distance of P from moon=D-x

$F_m$ =Force applied on rest mass due to m

$F_m$ =Force on rest mass due to mas M

$F_M=F_m$ because net force is equal to 0.

$F_m=F_M$

$\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}$

$\frac{m}{(D-x)^2}=\frac{M}{x^2}$

$\frac{x^2}{(D-x)^2}=\frac{M}{m}$

$\frac{x}{D-x}=\sqrt{\frac{M}{m}}$

Let $R=\sqrt{\frac{M}{m}}$

Then, $\frac{x}{D-x}=R$

$x=DR-xR$

$x+xR=DR$

$x(1+R)=DR$

$x=\frac{DR}{1+R}$

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x= $\frac{2}{3}D$

Net force is zero

$F_m=F_M$

$\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}$

$\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}$

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

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0

4 years ago

A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe

nataly862011 [7]

Answer:

$v_x = 11.78 m/s$

Explanation:

Velocity of the ship is given as

$v = 12 units$

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

$v_x = v cos\theta$

$v_x = 12 cos11$

$v_x = 11.78 m/s$

7 0

3 years ago

A satellite is in circular orbit around the earth. The orbit the satellite is at a height of 420 km above the earth's surface. F

astraxan [27]

Answer:

7661.06 m/s

Explanation:

R = Radius of Earth = $6.38\times 10^6\ m$

h = Distance from the Earth = 420000 m

G = Gravitational constant = $6.674\times 10^{-11} N m^2/kg^2$

M = Mass of Earth = $5.98\times 10^{24}\ kg$

$V=\sqrt{g{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{GM}{R^2}{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{6.674\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^6)^2}{\frac{(6.38\times 10^6)^2}{6.38\times 10^6 + 420000}}}\\\Rightarrow V=7661.06\ m/s$

The orbital speed of the satellite is 7661.06 m/s

6 0

4 years ago