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Lubov Fominskaja [6]

3 years ago

15

Please help!!!!!!!!!!

1 answer:

wlad13 [49]3 years ago

5 0

Answer:

C. Acceleration due to gravity

Explanation:

which is 9.8 m/s2 on Earth. The formula for calculating weight is F = m × 9.8 m/s2.

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A 70.0 kg person jumping down from a 3.00 meter wall hits the ground at 7.70 m/s. He hits the ground with stiff legs and stops i

shusha [124]

Answer:

$F_{net} = 24434.783\,N$

Explanation:

Let consider that the person hit the ground in the negative direction. The physical model for the impact is modelled after the Impact Theorem:

$m_{person}\cdot v + Imp = 0$

The impact is:

$Imp = - m_{person}\cdot v$

$Imp = - (70\,kg)\cdot (-7.70\,\frac{m}{s} )$

$Imp = 539\,N\cdot s$

The force of impact is derived as follows:

$F_{net} = \frac{Imp}{\Delta t}$

$F_{net} = \frac{539\,N\cdot s}{0.023\,s}$

$F_{net} = 24434.783\,N$

3 0

3 years ago

2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.

guapka [62]

Answer:

The specific kinetic energy of a mass is 0.8 kJ/kg

Explanation:

Given that,

Velocity = 40 m/s

Specific kinetic energy is the kinetic energy per unit mass.

We need to calculate the specific kinetic energy

Using formula of specific kinetic energy

$K.E=\dfrac{\dfrac{1}{2}mv^2}{m}$

$K.E=\dfrac{v^2}{2}$

Put the value into the formula

$K.E=\dfrac{40^2}{2}$

$K.E=800\ J/kg$

We know that,

1 kJ = 1000 J

or, 1J=0.001 KJ

The specific energy is

$K.E=800\times0.001$

$K.E=0.8\ kJ/kg$

Hence, The specific kinetic energy of a mass is 0.8 kJ/kg

6 0

3 years ago

What is the energy of a photon with a frequency of 1.7 × 1017 Hz? Planck’s constant is 6.63 × 10–34 J•s.

Karo-lina-s [1.5K]

The energy carried by a single photon is given by
$E=hf$
where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of $f=1.7 \cdot 10^{17} Hz$ , therefore its energy is
$E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J$

3 0

3 years ago

Read 2 more answers

Why are the orbits of planets only nearly circular and not perfectly circular?

Galina-37 [17]

Explanation:

this is my answer this is helpful for you

7 0

3 years ago

A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T f

Studentka2010 [4]

Answer:

The current in the wire under the influence of the force is 216.033 A

Solution:

According to the question:

Length of the wire, l = 0.676 m

$\theta = 57.7^{\circ}$

Magnetic field of the Earth, $B_{E} = 5.29\times 10^{- 5} T$

Forces experienced by the wire, $F_{m} = 6.53\times 10^{-3} N$

Also, we know that the force in a magnetic field is given by:

$F_{m} = IB_{E}lsin\theta$

$I = \frac{F_{m}}{B_{E}lsin\theta}$

$I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}$

I = 216.033 A

5 0

3 years ago