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madam [21]
1 year ago
9

how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?

Physics
1 answer:
Mars2501 [29]1 year ago
8 0

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

Explanation:

Typically, the mathematical formula for kinetic energy is as follows:

KE=\frac{1}{2} *m*v^{2}

K.E= kinetic energy

M=mass

V= speed

so,

KE=1/2*0.008*(420)*420

K.E=672.4J

 Therefore,

v*v=672.4/85

v*v=15.821

v=4.0m/s

Therefore

v=4.0m/s

What is Kinetic Energy?

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

To know more about Kinetic Energy? Visit to: -

brainly.com/question/22174271

#SPJ4

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A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
Temperature is a measure of_______ of the particles in an object.
BigorU [14]

Answer:

temperature is a measure of hot or cold or warm or foggy

Explanation:

4 0
3 years ago
Read 2 more answers
Formula: KE = 1/2 my?
9966 [12]

Answer:

22400 Joules

Explanation:

Apply the formula:

KE = 1/2 . 40 . 1120

KE = 20 . 1120

KE = 22400 Joules

3 0
3 years ago
With e in volts per meter and t in seconds. at t = 0, the field is upward. the plate area is 4. 3 × 10-2 m2. for t &gt; 0, what
kenny6666 [7]

The magnitude of the displacement current between the plates is    2.1*10^{-8} A

Given,

A=4.3*10^{-2}  m^{2}

E=(4.0*10^{5})-(6.0*10^{4}t)\\

i_{d}  =ϵ_{0} *\frac{dϕ_{E} }{dt } =ϵ_{0}A\frac{dE}{dt}

i_{d}=ϵ_{0}*A*\frac{d}{dt}(4.0*10^{5})-(6.0*10^{4}t)=-ϵ_{0} *A*6.0*10^{4}

= -(8.85*10^{-12})(4.0×10*^{-2})(6.0×10^{4})=-2.1*10^{-8} A

<h3>Current </h3>

An electrical charge carrier flow known as current often involves electrons or atoms lacking in electrons. The capital letter I is frequently used as a symbol for current. Amperes are the common unit and are denoted by the letter A. A coulomb of electrical charge moves past a certain place in one second as one ampere of current does. Franklin current or conventional current are terms used by physicists to describe how current flows from relatively positive to comparatively negative sites. Negatively charged electrons are the most prevalent charge carriers. They move in a somewhat good direction from relatively negative points.

With e in volts per meter and t in seconds. at t = 0, the field is upward. the plate area is 4. 3 × 10-2 m2. for t > 0, what is the magnitude of the displacement current between the plates?

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5 0
1 year ago
x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering
ValentinkaMS [17]

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

2d\sin\theta=n\lambda

For second order, n = 2

\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

4 0
3 years ago
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