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anygoal [31]
3 years ago
6

Which equation is used to calculate the weight of an object on a planet?

Physics
2 answers:
kondaur [170]3 years ago
8 0

W=mg

Explanation:

the weight of an object on a planet is given by the law:

W=mg

where

m is the mass of the object

g is the value of the gravitational acceleration on the planet

While the mass of the object, m, depends only on the properties of the object (so, it is the same on every planet), the weight W, depends on the value of g, which is different from planet to planet, therefore the weight of an object changes if we move it to another planet.

Kamila [148]3 years ago
6 0

In this case to find the weight of an object you must use the formula.

W = mg

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what is the average gravitational force of attraction between the earth and the sun? the earth averages a distance of about 150
Tanzania [10]

Answer:

B

Explanation:

Hhhhh

5 0
3 years ago
A elephant kicks a 5.0\,\text {kg}5.0kg5, point, 0, start text, k, g, end text stone with 150\,\text J150J150, start text, J, en
S_A_V [24]

The speed of the stone is 7.7 m/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically,

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the stone in this problem, we have:

K = 150 J is its kinetic energy

m = 5.0 kg is its mass

Re-arranging the equation for v, we find the speed of the stone:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(150)}{5.0}}=7.7 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
densk [106]

Answer:

0.143 m

Explanation:

The relationship between force applied on a string and stretching of the spring is given by Hooke's law:

F=kx

where

F is the force exerted on the spring

k is the spring constant of the spring

x is the stretching of the spring from its equilibrium position

In this problem, we have:

F = 20 N is the force applied on the spring

k = 140 N/m is the spring constant

Solving for x, we find how far the spring will stretch:

x=\frac{F}{k}=\frac{20}{140}=0.143 m

7 0
3 years ago
You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po
nataly862011 [7]

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

4 0
3 years ago
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