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2 years ago

6

How can we define force?

2 answers:

mars1129 [50]2 years ago

8 0

Answer:

There are 2 definitions chose the suitable one for ur question

Explanation:

1. strength or energy as an attribute of physical action or movement.

2. coercion or compulsion, especially with the use or threat of violence.

HOPE THIS HELPS

zimovet [89]2 years ago

4 0

Explanation:

A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.

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A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0

3 years ago

Please help, it's an exam ://

LenaWriter [7]

Answer:

Whats the question

Explanation:

7 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc

Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity $v_1$ in time $t_1$

$\frac{d}{2}=v_1t_1----1$

from B to C

distance traveled is 0.5 d with $v_2$ and $v_3$ velocity for half-half time

$\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2$

divide 1 and 2 we get

$\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}$

$\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}$

Now average velocity is given by

$v_{avg}=\frac{d}{t_1+t_2}$

taking $t_1$ common

$v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}$

$v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}$

$v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}$

6 0

3 years ago

A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5

Phoenix [80]

Answer:

The propulsion force at the moment of departure, is 49 N

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N

Therefore, the propulsion force at the moment of departure, is 49 N

7 0

3 years ago