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4 years ago

Define equilibrium and explain first condition of equilibrium. Pease make it fast...

Physics

1 answer:

Semenov [28]4 years ago

4 0

Answer:

First Condition of Equilibrium

For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero. Here we will discuss the first condition, that of zero net force. ... Fnet=0 F net = 0.

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A toy helicopter takes off and moves 6 m up and then 1 m back down. What

klio [65]

the answer is 5 m up AKA "B"

6 0

4 years ago

A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower

scoray [572]

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

$d = 50 m$

And that the initial speed of the ball is

$u=25 m/s$

at an angle of

$\theta=20^{\circ}$

So, the horizontal speed of the ball (which is constant during the entire motion) is

$u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s$

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

$t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s$

which is the time the ball spent in air.

8 0

3 years ago

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

$L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m$

The velocity of the wave in the string is given by :

$v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s$

So, the velocity of the waves in the string is 72.54 m/s.

3 0

4 years ago

Find u + (3v - w) <br> u =-4i-8j <br> v= -11i -3j <br> w=5i+7j

k0ka [10]

Answer:

gunrun

Explanation:

5 0

2 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago