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2 years ago

A 2,300-kg truck is traveling down a highway at 32 m/s. What is the kinetic energy of the truck?

Physics

1 answer:

kondor19780726 [428]2 years ago

6 0

m = mass of the truck = 23 00 kg

v = speed of the truck down the highway = 32 m/s

K = kinetic energy of the truck = ?

kinetic energy of the truck down the highway is given as

K = (0.5) m v²

inserting the values

K = (0.5) (2300) (32)²

K = (0.5) (2300) (1024)

K = (1150) (1024)

K = 1177600 J

hence the kinetic energy of the truck comes out to be 1177600 J

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These the flow of electrons (the current) and where some of the electrons' energy gets converted into heat. (Lessons 5.01-5.03)

liq [111]

Answer:

Conductors.

Explanation:

Conductors allow the flow of electrons i. e. the current and where some of the electrons energy gets converted into heat while some electrons energy converted into light. Conductors are the materials which has the ability to allow heat and electricity to flow through them. Metals such as gold, copper, Silver, Aluminum, Mercury, Steel, Iron and salty water etc are good conductors of heat and electricity.

3 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

3 years ago

What energy transfer will a stretched rubber band have when let go

GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

3 0

2 years ago

When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a

Nana76 [90]

Answer:

Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

Фi = ∫ E .dA = $q_{int}$ / ε₀

the Gaussian surface is a sphere so its area is

A = 4 π r²

the charge inside is

q_{int} = Q

we substitute

E 4π r² = Q /ε₀

E = 1 / 4πε₀ Q / r²

To calculate the flow on the two surfaces

* Sphere

Ф = E A

Ф = 1 / 4πε₀ Q / r² (4π r²)

Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

r = d / 2

We look for the diagonal with the Pythagorean theorem

d² = L² + L² = 2 L²

d = √2 L

we substitute

r = √2 / 2 L

r = L / √2

L = √2 r

now we can calculate the area of the cube that has 6 faces

A = 6 L²

A = 6 (√2 r)²

A = 12 r²

the flow is

Ф = E A

Ф = 1 / 4πε₀ Q/r² (12r²)

Ф_cubo = 3 /πε₀ Q

the relationship of these two flows is

Ф_cube /Ф_sphere = 3 /π

8 0

3 years ago

The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma

Oliga [24]

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

$p=3\ kg\times 1.5\ m/s$

p = 4.5 kg-m/s

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

5 0

3 years ago