All categories

3 years ago

A person walks a distance of 3.0 km due

Physics

1 answer:

Nataly [62]3 years ago

4 0

Answer:

Speed=1.6km/hr. I'm not sure about b

You might be interested in

Look at the v-t graph a remote-controlled toy car below. At t = 0.0 s, the car is located at +10.0cm. What is the magnitude of t

andrezito [222]

Well idk if this helps but the formula to solve acceleration is

a=F/m=(100kg)=1.0m/s 2

6 0

3 years ago

A ____ is one of the tiny dots of light that form a grid on your screen.

Degger [83]

Hi there

The answer is pixel

Hope this helps

8 0

3 years ago

Read 2 more answers

NEED ASAP!! A box of mass 10 kg requires 20 N to slide it across a surface. What is the weight of the box? What is the coefficie

kaheart [24]

Answer: 7

Explanation: 7 is the superior number

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

You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the

podryga [215]

The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4

3 0

3 years ago

Read 2 more answers