Help for physical science u4 limiting reactants

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t

Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

$10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]$

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

$v_{f}^{2} =v_{o}^{2}-2*a*x$

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

$0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]$

4 0

2 years ago

Science question: How do humans use the magnetic field for navigation?

I am Lyosha [343]

Answer:

Magnetoreception (also magnetoception) is a sense which allows an organism to detect a magnetic field to perceive direction, altitude, or location. This sensory modality is used by a range of animals for orientation and navigation, and as a method for animals to develop regional maps.

Explanation:

4 0

2 years ago

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

$a = \frac{v^2}{r}$

At constant speed, we will have;

$v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1$

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0

3 years ago

What do we call the distance labeled from A to B and what could we do to the note played on an instrument to change that distanc

Korvikt [17]

Correct answer choice is :

A) From A to B is known as the wavelength and changing the pitch of the note will change its length.

Explanation:

The amount or quantity of period within two things, points, lines, etc. the state or fact of existing separate in space, as of one thing from another; remoteness. a linear amount of space: Seven miles is a distance too great to walk in an hour. Distance is a scalar quantity describing the interval in two points. It is just the measure of the interval.

7 0

2 years ago

Read 2 more answers

A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje

Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

$\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m$

6 0

2 years ago