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3 years ago

What is the slope of a line that is perpendicular to the x-axis?

1 answer:

7 0

That slope is called "undefined". There's no number
that describes it.

If you take any two points on the line and try to use them
to calculate the slope, you'll wind up trying to divide by zero,
and that's not even allowed in arithmetic.

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A 45.0 kilogram boy is riding a 15.0-kilogram bicycle with a speed of 8.00 meters per second. What is the combined kinetic energ

svlad2 [7]

Answer:

1920Joules

Explanation:

The formula for calculating the kinetic energy of a body is expressed as;

KE = 1/2 mv²

m isthe mass

V is the speed

For the two masses, the combined KE is expressed as;

KE = 1/2(m1+m2)v²

KE = 1/2(45+15)(8)²

KE = 1/2 * 60 * 64

KE = 30 * 64

KE = 1920J

Hence the combined kinetic energy of the boy and the bicycle is 1920Joules

3 0

3 years ago

You use a pulley system to lift a car engine. You apply a force of 120n and the pulley pulls on the engine with a force of 1050n

Savatey [412]

Given,

Effort force = 120 N

Load force= 1050 N

Mechanical advantage of a pulley is given by the ratio of load force to the effort force.

$M.A=\frac{Load force}{Effort force}$

$=\frac{1050}{120}$

$=8.75$

Therefore, the mechanical advantage of the given pulley is 8.75.

5 0

3 years ago

Read 2 more answers

The specific heat of iron is 0.45j/gk. The amount of heat needed to raise the temperature of the 12 g of iron by exactly 15k?

labwork [276]

Answer:

81 J.

Explanation:

From the question given above, the following data were obtained:

Specific heat capacity (C) = 0.45 J/gK.

Temperature change (ΔT) = 15 K

Mass = 12 g

Heat required (Q) =?

The heat required to raise the temperature of iron can be obtained as illustrated below:

Q = MCΔT

Q = 12 × 0.45 × 15

Q = 81 J

Therefore, the heat required to raise the temperature of the iron is 81 J.

8 0

3 years ago

Example of the word médium ?

Ira Lisetskai [31]

I am the medium one that in the family

4 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago