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2 years ago

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A crash test dummy with a mass of 55kg is placed into a car traveling at 40m/s. During the first trial the dummy does not wear a

seatbelt and comes to rest in a time of 0.2 seconds. In the second trial, the dummy wears a seatbelt and comes to rest in a time of 0.5 seconds. Calculate the force experienced by the dummy with and without a seatbelt.

1 answer:

s2008m [1.1K]2 years ago

4 0

a. The force experienced by the dummy with a seatbelt is 11,000 Newton.

b. The force experienced by the dummy without a seatbelt is 4,400 Newton.

<u>Given the following data:</u>

Mass = 55 kg

Initial velocity = 0 m/s

Final velocity = 40 m/s

Time A = 0.2 seconds

Time B = 0.5 seconds

a. To calculate the force experienced by the dummy with a seatbelt:

<h3>Newton's Second Law of Motion</h3>

In order to solve for the force acting on the dummy, we would apply Newton's Second Law of Motion.

<u>Note:</u> The acceleration of an object is equal to the rate of change in velocity with respect to time.

Mathematically, Newton's Second Law of Motion is given by this formula;

$F = \frac{M(v\;-\;u)}{t}$

Substituting the given parameters into the formula, we have;

$F = \frac{55(40\;-\;0)}{0.2}\\\\F = \frac{55\times 40}{0.2}\\\\F = \frac{2200}{0.2}$

Force = 11,000 Newton.

b. To calculate the force experienced by the dummy without a seatbelt:

$F = \frac{55(40\;-\;0)}{0.5}\\\\F = \frac{55\times 40}{0.5}\\\\F = \frac{2200}{0.5}$

Force = 4,400 Newton.

Read more on force here: brainly.com/question/1121817

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levacccp [35]

Answer:

Because of the formula $E=mc^2$

Explanation:

In this problem we are describing two different processes:

Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

$E=\Delta mc^2$

where:

E is the energy released in the reaction

$\Delta m$ is the mass defect, the difference between the final total mass and the initial total mass

$c=3.0 \cdot 10^8 m/s$ is the speed of light

From the formula, we see that the factor $c^2$ is a very large number, therefore even if the mass defect $\Delta m$ is very small, nuclear fusion and nuclear fission release huge amounts of energy.

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3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

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K is the maximum kinetic energy of the photoelectron

In this problem, we have

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Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

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The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

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Potassium

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C. 4.9 eV

In this case, we have

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So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

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Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

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Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

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Therefore, the electric field strength inside the capacitor is 49880.77 N/C

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Bingel [31]

Circular Motion does occur on earth, so yes, true

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Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let $\overset \rightarrow {v_y}$ represent the velocity of the boat across the river in the direction, $\overset \rightarrow {d_y}$ , and let, $\overset \rightarrow {v_x}$ , represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = $\overset \rightarrow {v_y}$

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is $\overset \rightarrow {v_y}$ , such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

$\overset \rightarrow {v_y}$ = 3 m/s

$\overset \rightarrow {v_x}$ = 4 m/s

The width of the river, $w_y$ = 6 meters, we have;

The resultant velocity = $\sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s$

The direction, θ, is given as follows;

$\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}$

The length of the path of the boat, <em>l</em>, is given as follows;

$l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10$

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = $\dfrac{l}{v}$ , therefore;

$t = \dfrac{10}{5} = 2$

The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

$t = \dfrac{w_y}{v_y}$

Therefore;

$t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s$

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

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