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IRISSAK [1]
4 years ago
14

If you are in a spaceship that is sitting on the surface of a planet, you feel your weight. How does this compare to the weight

you feel in an accelerating spacecraft?
Physics
1 answer:
nordsb [41]4 years ago
8 0

Answer:

You will feel more weight if it is accelerating out of the planet.

You will feel less weight if it is accelerating towards the planet.

Explanation:

The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.

When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.

When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.

If the spaceship is free falling at the gravitational acceleration you experience a zero weight

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
Name one similarity between Scientific Method and<br> CER
sp2606 [1]

Answer:

a

Explanation:

6 0
2 years ago
FILL IN THE BLANK.<br><br> The __________ of a vector is represented by the length of the arrow.
melisa1 [442]

Answer:

magnitude

Explanation:

because

5 0
3 years ago
App182.studyisland.com/cfw/test/practice-session/a56do?CFID=14068
Alex

The amount of matter contained by an object is called mass.

A. mass

<u>Explanation:</u>

Mass is fundamentally a property of any physical amount and it is additionally the estimation of the resistance from the acceleration when force is applied on an object. The mass equals the quality of the gravitational force on a body.

Mass, in material science, the quantitative proportion of idleness, a crucial property of all matter. The greater the mass of a body, the littler the change created by an applied power. The mass of an object can be portrayed by its capacity to oppose a given power (we once in a while call this a body's inertial mass and subsequently mass is personally connected with the idea of inertia).

This is a straightforward result of Newton's second law where the power F, on a body is equivalent to the mass m, times the speeding up an, it encounters, ie:

F=ma or m=F/a

Mass of an object can not be zero but weight can be zero. The mass and weight of an object are different things.

8 0
3 years ago
Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
4 years ago
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