Two forces of 5N and 7N respectively act on an object. When will the resultant of the two vectors be at a maximum?

For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 7.45 105 N. What is the m

ehidna [41]

Answer:

7.45 10^5N.

Explanation:

according to newtons second law of motion;

$\sum F_x = ma_x\\F_{app} - F_r = ma_x$

Fapp is the applied force

Fr is the resistive force

m is the mass of the luxury

a is the acceleration

Since the huge luxury liner move with constant velocity, then acceleration is zero i.e a = 0. The equation becomes;

$F_{app} - F_f = m(0)\\F_{app} - F_f =0\\F_{app} = F_f$

This shows that the applied force will be equal to the resistive force if the velocity is constant.

Given Fr = 7.45 10^5 N therefore the resistive force will also be 7.45 10^5N.

Hence the magnitude of the resistive force exerted by the water on the cruise ship is 7.45 10^5N.

8 0

3 years ago

Essay about why people should not join a gang 300 word

Salsk061 [2.6K]

Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

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-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

-Most of the time when someone joins a gang a child is being separated from a parent and will face most aggression.

-This can also cause academic failure in students, from lack of education and probably a pretty bad background.

-For teens when joining a gang they will do things that aren't suited for their age (I would list things but I'm not trying to get myself reported if you know what I mean)

-You will also have a great risk of being imprisoned and facing bad consequences, and these consequences can even be having the death penalty.

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I hope this is good enough, sorry I didn't give you a full essay. I think this should help with that though.

6 0

2 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

Which form of Kepler’s third law can you use to relate the period T and radius r of a planet in our solar system as long as the

ser-zykov [4K]

T2=r In the form of Kepler's law that can use to relate the period T and radius of the planet in our solar systems

Explanation:

Kepler's third law:

Kepler's third law states that For all planets, the square of the orbital

period (T) of a planet is proportional to the cube of the average orbital radius (R).

In simple words T (square) is proportional to the R(cube) T²2 ∝1 R³3

T2 / R3 = constant = 4π ² /GM

where G = 6.67 x 10-11 N-m2 /kg2

M = mass of the foci body

8 0

4 years ago

A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a d

Irina-Kira [14]

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

$F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m$

Extension of the spring would be 0.040625 m

b) Work done in a spring

$W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J$

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J

4 0

3 years ago