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3 years ago

15

What is the relationship between two variables if the product of the variables is constant?

1 answer:

OverLord2011 [107]3 years ago

6 0

Answer:

If the product of two variables (x ⋅ y) is equal to a constant (k = x ⋅ y), then the two are said to be inversely proportional to each other with the proportionality constant k

Explanation:

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On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V

From here, we solve for mass:

m = d * V (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>

4 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity

stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

$V_{ox}$ = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = $V_{ox}$ t

80 = (v Cos60) (t)

t = 160/v eq-1

consider the motion in vertical direction :

Y = vertical displacement = 20 m

$V_{oy}$ = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = $V_{oy}$ t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0

3 years ago

If you walk 4mi north and then 2mi south what is your total displacement?

sergeinik [125]

Answer:

your total displacement is 2 miles north

Explanation:

8 0

2 years ago

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a

MrRissso [65]

Answer:

$h>\dfrac{5}{2}R$

Explanation:

Given that

Height = h

Radius = R

From energy conservation

$KE_A+U_A=KE_B+U_B$

At point B

The minimum speed to complete the the circle

$V_B=\sqrt{gR}\ m/s$

So the kinetic energy at point B

$KE_B=\dfrac{1}{2}mV^2$

$KE_B=\dfrac{1}{2}mgR$

$KE_A+U_A=KE_B+U_B$

$0+mgh=\dfrac{1}{2}mgR+2mgR$

Without falling off at the top (point B)

$0+mgh>\dfrac{1}{2}mgR+2mgR$

$mg(h-2R)>\dfrac{1}{2}mgR$

$g(h-2R)>\dfrac{1}{2}gR$

$h>\dfrac{5}{2}R$

6 0

4 years ago