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3 years ago

A physical science test book has a mass of 2.2 kg what is the weight on mars??? please help

Physics

1 answer:

Nastasia [14]3 years ago

4 0

Weight = (mass) x (gravity)
= (2.2 kg) x (3.7 m/s^2)

= 8.1 newtons. (about 1.84 pounds)

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Triangulation is the preferred method of calculation for all astronomical distances.

yan [13]

if it was possible to triangulate all objects it would give the most accurate answer. So yes, it is the prefered method. However as this is not the case other methods are used, less accurate but more practical

8 0

3 years ago

Read 2 more answers

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to p

katrin [286]

Answer:

Required charge $q=2.6\times 10^{9}C$ .

$n=1.622\times 10^{10}\ electrons$

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

$E =\frac{kq}{r^2}$

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = $\frac{25.0}{2}=12.5cm=0.125m$

thus, according to the given data

$1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}$

$q=\frac{0.125^2\times 1500}{9\times 10^{9}}$

Required charge $q=2.6\times 10^{9}C$ .

Now,

the number of electrons (n) required will be

$n=\frac{required\ charge}{charge\ of\ electron}$

$n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}$

$n=1.622\times 10^{10}\ electrons$

6 0

4 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance

trapecia [35]

Answer:

v_f = 6.92 x 10^(4) m/s

Explanation:

From conservation of energy,

E = (1/2)mv² - GmM/r

Where M is mass of sun

Thus,

E_i = E_f will give;

(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)

m will cancel out to give ;

(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)

Let's make v_f the subject;

v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]

G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²

Mass of sun is 1.9891 x 10^(30) kg

v_i = 2.1×10⁴ m/s

r_i = 2.5 × 10^(11) m

r_f = 4.9 × 10^(10) m

Plugging in all these values, we have;

v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12

v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]

v_f = √[(441000000) + (435.38 x 10^(7))

v_f = 6.92 x 10^(4) m/s

5 0

4 years ago

A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 23.8 s. Assuming constant angular acceleration, what is it

ser-zykov [4K]

Answer:

$0.053 rad/s^2$

Explanation:

0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s

Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be

$\alpha = \Delta \omega / \Delta t = \frac{0.4 \pi - 0}{23.8} = 0.053 rad/s^2$