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3 years ago

PLEASE HELP

Physics

2 answers:

Deffense [45]3 years ago

8 0

Answer: a physical change

kirill [66]3 years ago

6 0

Answer:

physical change

Explanation:

i got it right on the test

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In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

Sphinxa [80]

Answer: $\frac{V^{2}}{r}$

Explanation:

According to Newton's 2nd Law of motion the force $F$ is proportional to the mass $Fm$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, the equation for the Centripetal force is:

$F=\frac{mV^{2}}{r}$ (2)

Where:

$V$ is the velocity

$r$ is the radius of the circular motion

Making (1) and (2) equal:

$m.a=\frac{mV^{2}}{r}$ (3)

Hence:

$a=\frac{V^{2}}{r}$ This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0

3 years ago

Read 2 more answers

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?

slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

<em>applied force, F = 600 N</em>
<em>mass of the truck, m = 1,500 kg</em>
<em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

$F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s$

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

7 0

3 years ago

How long will it take for a boat traveling at 36 k/h to go 126 km?

brilliants [131]

3 hours 30 min i believe because 126/36 is 3.5 and 3.5 hours is 3 hours and 30 min. Brainliest pls

5 0

3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

3 years ago