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3 years ago

Which dog has the most kinetic energy

1 answer:

4 0

The formula to find the kinetic energy is:

Ek= 1/2 × m × v^2

1. Ek= 1/2×15×3^2
= 67.5 J

2.Ek= 1/2×8×4^2
=64 J

3.Ek= 1/2×12×5^2
= 150 J

4.Ek= 1/2×10×6^2
= 180 J

So the fourth dog has the most kinetic energy.

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Granite is uplifted by the movement of tectonic plates. It will most likely become _____. igneous rock sedimentary rock metamorp

pashok25 [27]

Answer:

The answer would be Igneous rock

Explanation:

3 0

3 years ago

Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati

jek_recluse [69]

Answer:

$a=4\ m/s^2$

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2$

So, the magnitude of centripetal acceleration is $4\ m/s^2$ .

5 0

2 years ago

A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling

Thepotemich [5.8K]

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

7 0

3 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

3 years ago

A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric

andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0

3 years ago