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nikdorinn [45]
3 years ago
14

2. What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2 m/s?

Physics
2 answers:
telo118 [61]3 years ago
6 0

Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

Dimas [21]3 years ago
3 0

Answer:

The kinetic energy of bowling ball is 14.52J

Explanation:

Given that the formula of kinetic energy is (1/2)mv². Then substitute the following value into the formula:

m = 7.26kg

v = 2m/s

k.e = (1/2)×7.26×2²

= 14.52J

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What is the net work doneon the object over the distance shown?
GuDViN [60]

A)F_0d

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

so

Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

A)F_0d

I hope this helps you

4 0
1 year ago
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10
marissa [1.9K]

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

7 0
3 years ago
How does reflection differ from refraction and diffraction?
shtirl [24]

Answer: Reflection is the only process in which the wave does not continue moving forward.

Explanation:

Reflection is a process in which the direction of the wave changes when it is exposed to a bounce off barrier. Refraction can be defined as the change in the direction of the wave when the wave passes through one medium to another. Diffraction is a process in which the direction of the wave changes when the wave passes through a particular opening near the barrier.

7 0
2 years ago
A planet orbits a start with the path shown below.
kvv77 [185]

PART a)

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So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

4 0
3 years ago
If an object is placed at the center of carvature of a convance mirror the image formed is called
Natasha_Volkova [10]

Answer:

When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.

3 0
2 years ago
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