All categories

3 years ago

A car drives 2 miles north, 4 miles west, 2 miles south and then drives 6 miles east. What is the car's distance?

Physics

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

Ok, so the car has traversed a total of 2 + 4 + 2 + 6 = 14 miles.

That's the car's distance! :)

For the displacement, you can draw out a diagram. I'll try my best to make one using a keyboard xDDD

- - - - |

| |

| - - - start - end!

So, this is the journey of the car: Going two up, 4 left, 2 down, and 6 right.

This ends up 2 to the right of the beginning!

That's a 2 mile displacement :)

Can you give me brainliest for my splendid diagram? xD

You might be interested in

Quel est le type de rayonnement produit par la radioactivité

Dmitriy789 [7]

Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:

-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons

-- "rayons bêta" ou "particules bêta" . . . flux d'électrons

-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée

8 0

3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

4 years ago

"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3

avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by $B=\frac{\mu _0i}{2\pi r}$

It is given that magnetic field is zero at some distance so

$\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}$

Here $i_1=3\ A \ and\ i_2=8\ A$

So $\frac{3}{x}=\frac{8}{24-x}=6.5454\ m$

3 0

3 years ago

A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of

grigory [225]

Answer:

$a=2.9\ m/sec^2$

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force $F_r$ acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

$\vec F_r=m\vec a$

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force $F_r$ is computed as the hypotenuse of a right triangle

$|F_r|=\sqrt{256^2+104^2}$