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N76 [4]
3 years ago
7

Hii please help i’ll give brainliest!!

Physics
1 answer:
weqwewe [10]3 years ago
4 0

Answer:

Hey buddy, it is D

Explanation: Just do the math, take 30 and then subtract 20 ok and then yo would have 10, then the 10is really the 30 and the 30 is pointing to the left so 10 newtons to the left and also your welcome

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A small car is pushing a large truck. They are speeding up.
Gnom [1K]

The force of the truck on the car is equal to the force of the car on the truck.

<h3>What is Newton's third law of motion?</h3>
  • Newton's third law of motion states that action and reaction are equal and opposite.

Let the mass of the small car = m

Let the mass of the truck = M

The force exerted by the small car is calculated as;

F₁ = ma

The force exerted by the large truck is calculated as;

F₂ = Ma

According to Newton's third law, the magnitude of the two forces are equal but opposite in direction.

|F₁| = |F₂|

Thus, we can conclude that, the force of the truck on the car is equal to the force of the car on the truck.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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2 years ago
A rocket experiences a constant force even as the amount of fuel in its fuel tanks decreases. What happens to the acceleration o
Elanso [62]
The correct answer is D
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3 years ago
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A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t
Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

\omega= 19 rev/s

we need to convert it into radiands per second. We know that

1 rev = 2 \pi rad

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

\theta= \omega t

where

\omega=119.3 rad

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

\theta=(119.3 rad/s)(5 s)=596.5 rad

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

\alpha = 1.6 rad/s^2

So the final angular speed will be given by

\omega_f = \omega_i + \alpha \Delta t

where

\omega_i = 119.3 rad/s is the initial angular speed

\alpha = 1.6 rad/s^2 is the angular acceleration

\Delta t = 15 s - 10 s = 5 s is the time interval

Solving the equation,

\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2

Substituting the numbers into the equation, we find

\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad

(e) 222 m

The instantaneous speed of the center of the wheel is given by

v_{CM} = \omega R (1)

where

\omega is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s

And so the displacement of the center of the wheel will be

d=v_{CM} t = (44.4 m/s)(5 s)=222 m

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In liquids, molecules particles are close together but move in random directions.

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Less than 7, Ph of 7 is neutral Acid Rain would be between 4 and 0. 
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