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2 years ago

5

A man is pushing a large box with a force of 44 newtons. He slides over a floor that has a sliding frictional force of 12 Newton

s. What is the acceleration?

1 answer:

adoni [48]2 years ago

7 0

I’m sorry I was not texting lol but 12 subduing and I got a phone call back to you later on for your birthday 17 wyd

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A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

$Torque = \vec{r} \times \vec{F}$

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

$Torque = \vec{r} \times \vec{F}$

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :

$Torque = rFsin(\theta)}$

$65 N m = (0.35m)Fsin(90°)}$

$F = \dfrac{65N m} {(0.35m)sin(90°)}$

$F = \dfrac{65N m} {(0.35m)sin(90°)}$

$F = \dfrac{1300} {7}N$

so the force is approximately 186 N.

4 0

4 years ago

If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil

Hoochie [10]

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

5 0

3 years ago

If an object looses electrons, then have been transferred

slava [35]

If<span> a neutral </span>object loses<span> some </span>electrons<span>, </span>then<span> it will possess more protons</span>

4 0

3 years ago

In the Bohr’s model of the hydrogen atom, the electron moves in a circular orbit of radius 5.041 × 10−11 m around the proton. As

Ilia_Sergeevich [38]

Answer:

The orbital speed of the electron is 2.296 x 10⁶ m/s

Explanation:

Given;

radius of the circular orbit, r = 5.041 × 10⁻¹¹ m

In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;

$V = \frac{nh}{2\pi mr}$

where;

h is Planck's constant

m is mass of electron

r is the radius of the circular orbit

n is the energy level of hydrogen in ground state

Substitute in these values and solve for V

$V = \frac{nh}{2\pi mr} = \frac{1*6.626*10^{-34}}{2\pi *9.11 *10^{-31}*5.041*10^{-11}}\\\\V = 2.296*10^6 \ m/s$

Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s

4 0

4 years ago

3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant

lions [1.4K]

The total distance traveled by the car at the given velocity and time is 900 m.

The given parameters:

<em>initial velocity of the car, u = 20 m/s</em>
<em>acceleration of the car, a = 12 m/s²</em>
<em>time of motion of the car, t = 20 s</em>
<em>final time = 30 s</em>
<em>final acceleration = 2 m/s²</em>

The final time of motion of car before coming to rest is calculated as follows;

$v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s$

The graph of the car's motion is in the image uploaded.

The total distance traveled by the car is calculated as follows;

$total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m$

Thus, the total distance traveled by the car at the given velocity and time is 900 m.

Learn more about velocity-time graph here: brainly.com/question/24874645

3 0

3 years ago