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3 years ago

According to Newton's third law of motion, if you push against a wall, the wall will __________.

2 answers:

8 0

It would be C as the law says "Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object."

Svetach [21]3 years ago

8 0

C push againt you with equal force.

jkhgfgd

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If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

AnnZ [28]

To solve this problem we will apply the concepts related to the centripetal Force and the Force given by weight and formulated in Newton's second law. Through the two expressions we can find the radius of curve made in the hand. To calculate the normal force, we will include the concepts of sum of forces to obtain the net force on the body at the top and bottom of the maneuver. The expression for centripetal force acting on the jet is

$F_c = \frac{mv^2}{r}$

According to Newton's second law, the net force acting on the jet is

F = ma

Here,

m = mass

a = acceleration

v = Velocity

r = Radius

PART A ) Equating the above two expression the equation for radius is

$\frac{mv^2}{r} = ma$

$r = \frac{v^2}{a}$

Replacing with our values we have that

$r = \frac{(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{7(9.8m/s^2)}$

$r = 1.462*10^3m$

PART B )

- The expression for effective weight of the pilot at the bottom of the circle is

$N = mg +\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)+\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 5408.87N$

Note that the normal reaction N is directed upwards and gravitational force mg is directed downwards. At the bottom of the circle, the centripetal force is directed upwards. So the centripetal force is obtained from the gravitational force and the normal reaction.

- The expression for effective weight of the pilot at the top of the circle is

$N = mg -\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)-\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 4056.47N$

Note that at the top of the circle the centripetal force is directed downwards. So the centripetal force is obtained from normal reaction and the gravitational force.

4 0

3 years ago

How much power is needed to lift the 200-N object to a height of 4 m in 4 s?

Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

3 0

3 years ago

IN WHAT CONDITION DO SOUND ECHO

DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0

2 years ago

How does friction affect the kinetic energy of an object?

Law Incorporation [45]

Answer: Friction also prevents an object from starting to move, such as a shoe placed on a ramp. When friction acts between two surfaces that are moving over each other, some kinetic energy is transformed into heat energy. Friction can sometimes be useful.

Explanation:

8 0

3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

3 years ago