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blsea [12.9K]
3 years ago
6

The weight of an astronout is 60kg on earth, find the wight of the same object on the planet where the gravitational attraction

has been reduced to 1/10 of the earths pull
Physics
1 answer:
Marianna [84]3 years ago
3 0

The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.

If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).

That's 588 Newtons on Earth, and 58.8 Newtons on a planet with 1/10 Earth's gravity.

The astronaut's mass of 60 kg goes with her, and doesn't depend on where she is.

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The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M \propto \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

6 0
3 years ago
Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward
lbvjy [14]

Answer:

Final velocity at the top of the ramp is 6.58m/s

Explanation

Check the attachment

4 0
3 years ago
at which plate boundary would you expect to see volcanoes A.divergent b.all of the above c.convergent D.transform
natulia [17]
A or d would be most likely it
7 0
3 years ago
A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degr
Solnce55 [7]

Answer:

7.7 km 26°

Explanation:

The total x component is:

x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87

The total y component is:

y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38

The magnitude is:

d = √(x² + y²)

d = 7.7 km

The direction is:

θ = atan(y/x)

θ = 26°

5 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
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