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zhannawk [14.2K]

3 years ago

11

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object?

1 answer:

hichkok12 [17]3 years ago

7 0

Hello!

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object ?

Data:

F (force) = 5 N

m (mass) = ?

a (acceleration) = 2 m/s²

Solving:

$F = m*a$

$5 = m*2$

$2\:m = 5$

$m = \dfrac{5}{2}$

$\boxed{\boxed{m = 2.5\:kg}}\end{array}}\qquad\quad\checkmark$

Answer:

2.5 kg

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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Explanation:

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So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;

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Therefore we need about 39312 turns of wire. Considering that each turn must have a length of $\pi\,D$ , where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:

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3 years ago

A pickup truck, initially stationary at a stop light, accelerates at a rate of 1.60m/s2 for 14.0 s. The truck then cruises at co

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Answer:

The total distance traveled by the truck is 1797 m

Explanation:

Hi there!

The equation of position and velocity of an object moving in a straight line with constant acceleration are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the truck at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time

v = velocity at time t.

Let's calculate the position of the truck after the first 14.0 s:

x = x0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference at the first stop light, the initial position, x0, is zero. Since the truck starts from rest, v0 = 0. So, the equation of position will be:

x = 1/2 · a · t²

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x = 157 m

Then, the truck travels with constant speed (a = 0) for 70.0 s. The equation of position will be:

x = x0 + v · t

In this case, let's consider the initial position as the the position where the car is after 14.0 s (157 m from the stop light). The velocity is the velocity reached after the 14.0 s of acceleration. Let's calculate it with the equation of velocity:

v = v0 + a · t (v0 = 0)

v = 1.60 m/s² · 14.0 s

v = 22.4 m/s

So, the position will be:

x = 157 m + 22.4 m/s · 70.0 s

x = 1725 m

Now, the truck slows down with an acceleration of 3.50 m/s² until it stops (until its velocity is zero). Let's calculate the time at which the velocity of the truck is zero:

v = v0 + a · t

0 = 22.4 m/s - 3.50 m/s² · t

-22.4 m/s / -3.50 m/s² = t

t = 6.4 s

Now let's calculate the position of the truck after that time considering the initial position as the position at which the truck was after the 70.0 s traveling at constant speed (1725 m from the stop light):

x = x0 + v0 · t + 1/2 · a · t²

x = 1725 m + 22.4 m/s · 6.4 s + 1/2 · (-3.50 m/s²) · (6.4 s)²

x = 1797 m

The total distance traveled by the truck is 1797 m

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3 years ago