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3 years ago

Solve the equation. 5x-5=10x+2

Physics

2 answers:

Veronika [31]3 years ago

8 0

Answer:

x = - 1.4

Explanation:

-5=10x+2-5x (subtract 5x from both sides)

-5=5x+2 (simplify)

-5-2=5x (subtract 2 from both sides)

-7=5x (simplify)

x=-7/5 (divide both sides by 5)

x=-1.4 (simplify)

i would really appreciate getting a brainliest. anyways i hope this helped and have a great rest of your day/night!! :)

murzikaleks [220]3 years ago

5 0

5x-5=10x+2

5x-5x-5=10x-5x+2

-5-2=5x+2-2

-7/5=5x/5

x=-7/5

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In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller i

kotykmax [81]

The maximum angular speed of the roller is 3.47rad/a,

the maximum tangential speed of the point an the rim of the roller is 1.47m/s,

the time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation is t = 2.78s,

5.4 rotation has the roller turned between t=0 and the time found in part c

<h3>What does tangential speed refer to?</h3>

Any item moving along a circular path has a linear component to its speed called tangential velocity. An object's velocity is always pointed tangentially when it travels in a circle at a distance r from the center. The word for this is tangential velocity.

To calculate the tangential speed, divide the circumference by the time required to complete one spin. For instance, if it takes 12 seconds to complete one rotation, divide 18.84 by 12 to get 1.57 feet per second as the tangential velocity.

(a) angular speed w = dθ/dt = 5t - 1.8t^2

dw/dt = 5 - 3.6t = 0 for max w

so max w occurs at t = 5/3.6 s = 1.39s

so w max = 5*1.39 - 1.8*(1.39)^2 = 3.47 rad/s

(b) tangential speed v = r*w

r = D/2 = 0.5m

so v = 0.5*w = 1.74 m/s

so t = 5/1.8s = 2.78s (or t = 0 invalid)

After t = 2.87s, w is negative (starts reversing direction of rotation)

Driving force would actually have to be removed some time before t=2.78s because the roller can't stop instantaneously, but insufficient info to calculate this.

(d) Up to t = 2.78s, θ = 2.5*(2.78)^2 - 0.6*(2.78)^3 rad = 33.95 rad = 5.40 rotations

Therefore,

A) 3.47rad/a

B) 1.47m/s

C) t = 2.78s

D) 5.4 rotation

To learn more about tangential speed, refer to:

brainly.com/question/19660334

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4 0

2 years ago

Which of the following is true of alternating current? Select all that apply.

Nitella [24]

Answer:

The correct option is (C).

Explanation:

Electric current can be direct or alternating.

The direct current flows only in one direction. It is unidirectional.

The alternating current or an C current reverse its direction periodically.

So, the correct statement regarding the alternating current is (c) i.e. Electrons reverse direction periodically.

7 0

3 years ago

Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022k

Taya2010 [7]

Answer: $g=1.97 m/s^2$

Explanation:

Given

mass of Jupiter is $M=4.9\times 10^{22} kg$

Density of Jupiter is same as Earth

$density\ of\ Earth=density\ of\ Jupiter=5510 kg/m^3$

$mass=volume\times density$

considering Jupiter to be sphere of radius r

$M=\frac{4}{3}\times \pi r^3\times \rho$

$r^3=\frac{3M}{\rho \times 4\pi}$

$r^3=\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi}$

$r=(\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi})^{\frac{1}{3}}$

$r=1.28\times 10^6 m$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

$g=\frac{6.67\times 10^{-11}\times 4.9\times 10^{22}}{(1.28\times 10^6)^2}$

$g=1.97 m/s^2$

3 0

3 years ago

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = $\sqrt{x^{2} + y^{2} }$

Then, the module of c will be:

module of c = $\sqrt{(xa + xb)^{2} + (ya + yb)^{2}}$ = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = $\sqrt{xb^{2} + yb^{2} }$ = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = $\sqrt{(2342 mi)^{2} + (234.4 mi)^{2} }$ = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0

3 years ago

Anything that has mass and occupies space is defined as.

forsale [732]

Answer:

matter is the correct answer.

8 0

2 years ago