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4 years ago

A 54-kg jogger is running at a rate of 3m/s.What is the kinetic energy f the jogger

Physics

2 answers:

spin [16.1K]4 years ago

8 0

Formula:
KE = 1/2mv^2
KE: Kinetic Energy m:mass v:velocity

Solution:
KE = 1/2 × 54 × (3)^2
KE = 1/2 × 54 × 9
KE = 243J

leonid [27]4 years ago

6 0

KE=1/2mv²
KE=1/2*54*3²
KE=243 J

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Which type of energy is stored in fuels such as gasoline and batteries

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Answer:

Chemical Energy

Explanation:

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3 years ago

The law of reflection states that if the angle of incidence is 39 degrees, the angle of reflection is ___ degrees.

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3 years ago

When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

Nesterboy [21]

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

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7 0

4 years ago

Read 2 more answers

The change in the momentum of an object is represented by the following formula:

Ede4ka [16]

Hi there!

Recall that:

Change in momentum = mass × change in velocity

Or:

Δp = mΔv = m(vf - vi)

Plug in the given values. We can assign east to be positive and west to be negative in this instance (Velocity is a vector with direction).

Thus:

Δp = (1)(-21 - 10) = -31 kgm/s OR 31 kgm/s WEST.

The correct answer is B.

Change in momentum is EQUIVALENT to the quantity of IMPULSE.

The correct answer is H.

6 0

2 years ago

A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t

jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}$

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

$\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})$

$\text { Work }=-15.03 \mathrm{J}$

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is $\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

$\mathrm{F}=13 \times 9.8$

F = 127.4N

The force exerted on the object is 127.4N.

4 0

3 years ago