Ask question

Ask question

All categories

1 year ago

10

Solve ~

13x - 26 + 39 = 0" align="absmiddle" class="latex-formula">

ty! :)

2 answers:

Natalka [10]1 year ago

4 0

Explanation:

13x - 26 + 39 = 0

13x=26-39

13x=-13

x= -1

liq [111]1 year ago

3 0

$13x-26+39=0\\\\\implies 13x +13 = 0\\\\\implies 13x = -13\\\\\implies x = -\dfrac{13}{13}\\\\\implies x = -1$

You might be interested in

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago

Two objects of different masses are sitting on different balance scales. Object A has a greater mass than object B. How will the

kodGreya [7K]

As long as they're both on the same planet, the greater mass always has the greater weight. In this question, Object-A has the greater mass, so it weighs more that Object-B does.

5 0

2 years ago

PLEASE HELP!!!!!!! A student is trying to demonstrate static electricity, so they rub two identical balloons with a neutral rabb

vlada-n [284]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

4 0

2 years ago

Gravitational energy can be negative?

AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

6 0

3 years ago