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AnnZ [28]
3 years ago
6

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

traveling with constant velocity.
Physics
1 answer:
artcher [175]3 years ago
5 0

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

0

Explanation:

The weight is defined as:

W = mg (1)

Where m is the mass and g is the gravity

a) On Earth g = 9.8m/s^{2}

Then, equation 1 can be used:

W = (50Kg)(9.8m/s^{2})

W = 490Kg.m/s^{2}

but 1N = Kg.m/s^{2}

W = 490N

Hence, the weight of the astronaut on Earth is 490N

b) On the Moon g = 1.7m/s^{2}

W = (50Kg)(1.7m/s^{2})

W = 85N

Hence, the weight of the astronaut on the Moon is 85N

c) On Mars g = 3.7m/s^{2}

W = (50Kg)(3.7m/s^{2})

W = 185N

Hence, the weight of the astronaut on Mars is 185N

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

a = \frac{v_{f} - v_{i}}{t}

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.      

W = (50kg)(0)

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

 

W = 0

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A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are
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Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

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Initial speed of the car, u = 1 m/s

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Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)

v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N

The force required to push to stop the car is 288.67 N

3 0
3 years ago
Why do nonmetals have a negative oxidation number
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A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i
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<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V =  I x R                -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R               -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance.  Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

6 0
3 years ago
Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical
gizmo_the_mogwai [7]

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

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