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2 years ago

8

A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000

kilograms?

1 answer:

Zigmanuir [339]2 years ago

8 0

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1.

</span>

<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>

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23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

PSYCHO15rus [73]

Answer:

a) (95.4 i^ + 282.6 j^) N , b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

sin θ = Fjy / Fj

cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

T = 90 -32 = 58º

Fjy = Fj sin 58

Fjx = FJ cos 58

Fjx = 180 cos 58 = 95.4 N

Fjy = 180 sin 58 = 152.6 N

Andrea's force is

Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

Fx = Fjx

Fx = 95.4 N

Axis y

Fy = Fjy + Fa

Fy = 152.6 + 130

Fy = 282.6 N

F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

F² = Fx² + Fy²

F = √ (95.4² + 282.6²)

F = √ (88963)

F = 298.27 N

tan θ = Fy / Fx

θ = tan-1 (282.6 / 95.4)

θ = tan-1 (2,962)

θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

F' = 298.27 N

θ' = 180 + 71.3

θ = 251.4º

4 0

2 years ago

Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh

ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0

2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t

Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s

when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

6 0

2 years ago

Read 3 more answers

A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.

charle [14.2K]

Answer:

(a)F= 3.83 * 10^3 N

(b)Altitude=8.20 * 10^5 m

Explanation:

On the launchpad weight = gravitational force between earth and satellite.

W = GMm/R²

where R is the earth radius.

Re-arranging:

WR² / GM = m

m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg

The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:

Fc = mω²r

where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.

ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s

When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:

Fc = GMm/r²

mω²r = GMm / r²

ω²r = GM / r²

r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²

r³ = 3.612 * 10^20

r = 7.12 * 10^6 m

(a) F = GMm/r²

F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²

F= 3.83 * 10^3 N

(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m

4 0

2 years ago