All categories

3 years ago

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m? (The acceleration due to gravity is 9.8 m/s2.)

Physics

2 answers:

V125BC [204]3 years ago

7 0

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?
Answer:

85 Joule (approx)

Explanation:

Potential energy at highest point

P.E = mgh = 7.2 kg × 9.8 m/s2×1.2 m≈85 Joule

This is the kinetic energy required for dog to jump a height of 1.2 m.

hope this helps!

Tasya [4]3 years ago

6 0

Answer:

Kinetic energy of dog is 84.67 Joules

Explanation:

Given that,

Mass of the dog, m = 7.2 kg

We have to find the kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m. We know that the energy of a system remains conversed. At highest point there is only potential energy. So, while taking a vertical jump the potential energy gets converted to potential energy.

So, $E_k=mgh$

here, h = 1.2 m

$E_k=7.2\times 9.8\times 1.2$

$E_k=84.67\ Joules$

Hence, the kinetic energy needed to make a vertical jump is 84.67 Joules.

You might be interested in

A 1400-kg car is moving at a speed of 35 m/s. How much kinetic energy does the car have (in Joules)?

Gekata [30.6K]

Answer:

857.5Kjoules

Explanation:

Using K. E=1/2 Mv^2

=1/2x1400x35x35

=857500joules

=857.5KJ

3 0

3 years ago

2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:

Alekssandra [29.7K]

Answer:

a=24.5 b=9.5

Explanation:

7 0

3 years ago

HEEEEELLLPPPPPP!!!!!!!!!!! PPPPPPPPPPPLLLEAASSEE!!!

o-na [289]

Answer:

The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart (repel).

4 0

3 years ago

if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,

ipn [44]

False.

The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
$U=mgh=(0.2 kg)(9.81 m/s^2)(0.3 m)=0.6 J$

This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.

7 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago