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3 years ago

9

If an astronaut has a mass of 80 kg when measured on Earth. What is the astronaut’s weight on Mars where the acceleration due to

gravity on Mars is 3.8 m/s/s?
Question 6 options:

Zero Newtons because objects in space are weightless

784 N

304 N

80 N

1 answer:

AlekseyPX3 years ago

6 0

Answer:

784 ni

Explanation:

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

4 years ago

What are 2 ways organisms would be affected by water pollution?

Ymorist [56]

Not having clean water to drink or bath. Also some organizations that live in the water might die because the water is polluted and that would be toxic form them

7 0

4 years ago

If a reaction starts with 30 grams how many should it end with ?

zysi [14]

30grams

Explanation:

If a reaction starts with 30grams then the reaction should end with 30grams.

This in conformity with the law of conservation of mass.

The law states that "in an isolated system, mass is neither created nor destroyed during chemical transformation".
Mass is the quantity of matter contained in a substance.
In chemical reactions, the mass of reactants must always be the same with the mass of the product baring any loss.
In an isolated system, there is no exchange of energy and mass.
Chemical systems are usually treated as isolated systems in which mass is conserved.

Learn more:

Chemical laws brainly.com/question/5896850

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4 0

3 years ago

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.

rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = $\frac{1}{2}$ $m_{p}$ v²

v = √( 2K / $m_{p}$ )

lets relate the cross-sectional area A of the beam to its diameter D;

A = $\frac{1}{4}$ πD²

now, we substitute for v and A

n = I / $\frac{1}{4}$ πeD² ×√( 2K / $m_{p}$ )

n = 4I/π eD² × √( $m_{p}$ / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0

3 years ago

Two small identical metal spheres, A and B, on insulated stands, are each given a charge of +2.0 x 10-5 coulomb. The distance be

SpyIntel [72]

Answer:

a) 90 N

Explanation:

Coulomb's law: states that the electric force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

It is represented mathematically as

F = 1/4πε₀(q₁q₂)/d²......................... Equation 1

Where F = Electrostatic force, q₁ = charge on the first metal sphere, q₂ = charge on the second metal sphere, d = distance between the spheres, 1/4πε₀ = proportionality constant.

<em>Given: q₁ = +2.0×10⁻⁵ C, q₂ = 2.0×10⁻⁵, d = 2.0×10⁻¹ m</em>

<em>Constant : 1/4πε₀ = 9×10⁹Nm²/C²</em>

<em>Substituting these values into equation 1,</em>

<em>F = (9×10⁹×2.0×10⁻⁵×2.0×10⁻⁵)/(2.0×10⁻¹)²</em>

<em>F = (9×4×10⁻¹)/4×10⁻²</em>

<em>F = 90 N</em>

Therefore the electrostatic force is = 90 N

The right option is a) 90 N

5 0

3 years ago