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3 years ago

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The average lifetime of a poodle dog is 13.0 years. How fast is it traveling, u, relative to an observer who measures the averag

e lifetime to be 50.0 years? 4

1 answer:

bogdanovich [222]3 years ago

6 0

Answer:

The speed is 0.97 c.

Explanation:

Given that,

Dilated time t'= 50.0 years

Rest time t = 13.0 years

We need to calculate the speed

Using formula of time dilation

$t'=\dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Where, t' = change in time

t = rest time

v = velocity

c = speed of light

Put the value into the formula

$50.0=\dfrac{13.0}{\sqrt{1-\dfrac{v^2}{(3\times10^{8})^2}}}$

$v^2=\dfrac{(13)^2\times(c)^2-(c)^2\times50^2}{50^2}$

$v^2= 0.9324c^2$

$v=0.97c$

Hence, The speed is 0.97 c.

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We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (

kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I = 476 \ N \cdot s$

b

$I_1 = 14.21 \ N\cdot s$

c

$F = 20300 \ N$

Explanation:

Considering the first question

From the question we are told that

The force produced is $F = 3400 \ N$

The duration of the punch is $t = 0.14 \ s$

Generally the impulse delivered is mathematically represented as

$I = F * t$

=> $I = 3400 * 0.14$

=> $I = 476 \ N \cdot s$

Considering the second question

The approaching velocity of the ball is $v_b = 45 \ m/s$

The leaving velocity of the ball is $v_l = -53 \ m/s$

The mass of the ball is $m_b = 0.145 \ kg$

Generally the magnitude of the impulse delivered is mathematically represented as

$I_1 = m* v_b - m * v_l$

=> $I_1 = [0.145 * 45] - [0.145 * -53]$

=> $I_1 = 14.21 \ N\cdot s$

Considering the third question

The duration of the impact of the bat is $t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s$

Generally the average force exerted by the bat is mathematically represented as

$F = \frac{I_1}{t_1}$

=> $F = \frac{14.21 }{0.7 *10^{-3}}$

=> $F = 20300 \ N$

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4 years ago

The forces acting on a sailboat are 390n north and 180n east. if the boat (including crew) has a mass of 270 kg, what are the ma

patriot [66]

F=ma so a=F/m

ax=180/270=0.67m/s^2
ay=390/270=1.44m/s^2

Magnitude = sqrt((0.67^2)+(1.44^2))=1.59m/s^2

Direction- Tan(x)=0.67/1.44=0.47 Tan^-1(x)=25 degrees

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3 years ago

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Which of the following transfers energy in the form of vibrating electric and magnetic fields? A. An electromagnet B. A generato

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Electromagnetic wave

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3 years ago

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Arte-miy333 [17]

Answer:

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3 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago