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Dafna1 [17]
3 years ago
10

Can anyone please help me with answering this question? : An accelerating (speeding up) body has a Net Force acting in the direc

tion of motion. True or False? Why? ​
Physics
1 answer:
Katarina [22]3 years ago
7 0
That's true.

Netwon's second law states that the resultant of the forces F acting on a body is equal to the product between its mass m and its acceleration a:
\sum F = ma
This means that if the net force acting on an object is different from zero (term on the left), than the acceleration of the object (term on the right) must be different from zero as well, and therefore the body is accelerating.

In particular, both F and a in the equation are vectors: this means that if the acceleration is positive, F and a have the same direction. In this problem, the acceleration is positive (because the object is speeding up), therefore the force and the acceleration have same direction.
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A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,
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Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

     θ = 90°,

work done = F s cos 90°        ∵ cos 90° = 0

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What is the physical energy that stimulates sight?
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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
a pulley of diameter 15.0 cm is driven by a motor that revolves at 10 rpm. the pulley drives a 2nd pulley with diameter 10.0 cm.
suter [353]
6.67prm

diam1/diam2 = rpm1/rpm2
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use the pendulum equation to calcuate the period of a 1.50 pendulum. Remeber that the vaule of "g" is 9.81 m/s² please help ​
kati45 [8]

Answer:

L = 3.51

Explanation:

Pendulum equation is T = 2pi\sqrt{L/9.81}

T = 1.5 and we are solving for L

1.5=2\pi \sqrt{L/9.81}

square both sides to get 2.25 = 2\pi ( L/9.81)

multiply both sides by 9.81 then divide by 2 and 3.14 as a substitue for pi. The answer should be about 3.51 in length

L = 3.51

If this helps, mark me brainliest pls

3 0
3 years ago
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