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2 years ago

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What is the difference between real and apparent weightlessness?

1 answer:

MrRissso [65]2 years ago

4 0

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

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Suppose you hit a 0.058-kg tennis ball so that the ball then moves with an acceleration of 10 m/s2. If you were to hit a basketb

stiks02 [169]

Answer:

1 m/s²

Explanation:

Force = mass × acceleration

F = ma ............ Equation 1

Where F = force, m = mass, a = acceleration.

Given: m = 0.058 kg, a = 10 m/s²

Substitute into equation 1

F = 0.058(10)

F = 0.58 N.

If the same force was used to hit the baseball,

F = m'a

a = F/m'.............. Equation 2

Where M' = mass of the baseball.

Given: F = 0.58 N, m' = 0.58 kg.

Substitute into equation 2

a = 0.58/0.58

a = 1 m/s²

8 0

3 years ago

Read 2 more answers

By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)

Fofino [41]

Gravity on the surface of sun is given as

$g = \frac{GM}{R^2}$

here we know that

$M = 1.98 \times 10^{30} kg$

$R = 6.95 \times 10^8 m$

now we will have

$g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}$

$g = 273.4 m/s^2$

now we need to find the ratio of weight on surface of sun and on surface of Earth

$R = \frac{mg_{sun}}{mg_{earth}}$

$R = \frac{273.4}{9.8} = 28 times$

so weight will increase by 28 times

6 0

3 years ago

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g ( g=9.81 m/s² )

F=ma

= (1510)*( 0.75*9.81)

= 11109.825 N

4 0

3 years ago

Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos

Korvikt [17]

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 $10^{3/2}$ / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

4 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

3 0

3 years ago