A 3 Volt battery is connected in series to three resistors: 4, 6, and 2

3. Differentiate between: (a) area and volume

Margarita [4]

Answer:

area is 2d volume is 3d

Explanation:

Area refers to the size of two-dimensional surface. Volume refers to the size of a three-dimensional space.

8 0

3 years ago

If Bill threw a ball straight up on the Moon (g=1.6 m/s2) with a starting velocity of 22m/s from a cliff and it fell past him an

kap26 [50]

Answer:

Write two important of physical state

5 0

2 years ago

A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini

finlep [7]

<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁, object 2 be u₂

Final velocity object 1 be v₁, object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

Initial momentum = Final momentum

24 = 3v₁ + 6M

v₁ + 2M = 8

v₁ = 8 - 2M

Initial kinetic energy = Final kinetic energy

96 = 1.5 v₁² + 18 M

v₁² + 12 M = 64

Substituting v₁ = 8 - 2M

(8 - 2M)² + 12 M = 64

64 - 32M + 4M² + 12 M = 64

4M² = 20 M

M = 5 kg

Option 2 is the correct answer.

6 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago

A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m

vampirchik [111]

Kinetic energy, KE, is modeled by the formula $KE = \frac{1}{2}mv^2$ , where m is the mass in kg and v is the velocity in m/s.

In this scenario, mass and one-half are constant but the velocity changes.

You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.

7 0

3 years ago