All categories

2 years ago

What is the acceleration of the box shown below.

Physics

2 answers:

-BARSIC- [3]2 years ago

4 0

Explanation:

Fnet= Fapp - F friction

=50-30

=20 N

Acceleration= force/ mass

=20/20

=1.0 m/s²

olga nikolaevna [1]2 years ago

3 0

Answer:

D) 1.0m/s^2

Explanation:

Force = 50 - 30 = 20

Acceleration = 20/20

= 1

You might be interested in

5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the

Darina [25.2K]

The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration ( $g=9.8 m/s^2$ towards the ground). Therefore, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

$a=-g=-9.8 m/s^2$ is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

$s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

3 0

3 years ago

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

6 0

4 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago

A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). The line runs i

Lelu [443]

Answer:

The first one is 3 m/s

The second one is 2 m/s

Explanation:

8 0

3 years ago

Read 2 more answers

What are the five plants you can see from earth without a telescope?

nevsk [136]

The five planets that you can see from Earth without a telescope are Mercury, Venus, Mars, Jupiter and Saturn.

3 0

3 years ago