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dalvyx [7]
3 years ago
10

A puppy weighing 3 kilograms races through the dog park. if she slows from a speed of 2 meters/second to 1 meter/second, what wi

ll happen to her kinetic energy? use the kinetic energy equation.
a.her kinetic energy decreases to 1.5 j
b.her kinetic energy increases to 1.5 j
c.her kinetic energy remains unchanged
d.her kinetic energy decreases to 1 j.
Physics
2 answers:
Ne4ueva [31]3 years ago
6 0
Your answer is A. If velocity decreases, your kinetic energy would also decrease.
sdas [7]3 years ago
5 0

Answer:The correct answer is option (a).

Explanation:

Mass of the puppy,m = 3 kg

Kinetic energy of the puppy when it was running with speed of 2 m/s:

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 3kg\times (2 m/s)^2=6 Joules

Kinetic energy of the puppy when it was running with speed of 1 m/s:

K.E'=\frac{1}{2}mv^2=\frac{1}{2}\times 3kg\times (1 m/s)^2=1.5 Joules

When the puppy was running at the speed f 1 m/s her kinetic energy reduced to 1.5 Joules.

Hence, correct answer is option (a).

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Helpp
frosja888 [35]
B, refraction I believe.
5 0
3 years ago
A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
4 years ago
A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its
sdas [7]
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

           s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² -  u x 2 - 0.5 x g x 2²

           s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² -  u x 4 - 0.5 x g x 4²

           s₅ = u - 4.5 g    

That is

           u - 2.5 g = u - 4.5 g + 3

             2 g = 3

                g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0
3 years ago
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
What does the term calorie refer to
OLga [1]
Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))

~Top
6 0
4 years ago
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