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3 years ago

The number of significant figures in the measurement 4.300×10^5 km are

Physics

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

Explanation

Any numbers in scientific notation are considered significant. For example, 4.300 x 10-4 has 4 significant figures.

Answer From Gauth Math

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

4 years ago

Which of the following are neutral particles?<br> Protons<br> Neutrons<br> Electrons<br> Atoms

emmasim [6.3K]

Answer:

i believe its an atom

Explanation:

7 0

3 years ago

Read 2 more answers

A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 16 km/s. To save the planet, astronauts str

blondinia [14]

Answer:

a. t = 69.4 hr = 2.89 days

b. theta = 91.67*10^-3 degrees

c. deflection_angle = 0.134 degrees

Explanation:

a).

The asteroid impacts the earth in t

t = x/v = (4.0*10^6 km)/(16 km/sec)

t = 2.5 * 10^5 sec

t = 69.4 hr = 2.89 days

b).

tan(theta) = 6400 km/(4.0*10^6 km)

tan(theta) = 1.6*10^-3

theta = arctan(1.6*10^-3)

theta = 1.6*10^-3 radians (for small angles, tan(theta) ~= theta)

theta = 91.67*10^-3 degrees

c).

v_minimum = 6400 km/(2.5 * 10^5 sec)

v_minimum = 25.6 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5.0*10^9 N = 4.0*10^10 kg * a

a = (5.0*10^9 N)/(4.0*10^10 kg)

a = 0.125 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 0.125 m/s^2 * 300 s

v_transverse = 37.5 m/s = 37.5*10^-3 km/s

tan(deflection_angle) = v_transverse/(20 km/s)

tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3

deflection_angle = arctan(2.34*10^-3)

deflection_angle = 2.34*10^-3 radians = 0.134 degrees

v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)

(note that the right hand side if this inequality is tan(theta) calculated above)

v_transverse > 23.704 m/

8 0

3 years ago

A 62.5 kg carpenter at a construction site plans to swing in a circular arc from one roof top to an adjacent roof at the end of

denis23 [38]

Answer:

v = 9.45 m/s

Explanation:

given,

mass of the carpenter = 62.5 Kg

length of rope = 13.4 m

Capable of exerting force = 1034 N

centripetal force acting on the body

$F = \dfrac{mv^2}{r}$

$F = \dfrac{62.5\times v^2}{13.4}$

F =4.664 v² N

Gravitational force on her =

F = m g

F = 62.5 x 9.81

F = 613.125 N

now,

4.664 v² + 613.125 = 1034

4.664 v² = 420.875

v² = 90.24

v = 9.45 m/s

Maximum speed which she can tolerate = v = 9.45 m/s

7 0

3 years ago

An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 si

spayn [35]

Answer:

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

$r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}$

$r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)$

r = 10

5 0

3 years ago

The number of significant figures in the measurement 4.300×10^5 km are​

The number of significant figures in the measurement 4.300×10^5 km are