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iVinArrow [24]
2 years ago
9

A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational

field strength of Planet X?
A. The gravitational field strength of Planet X is mg.

B. The gravitational field strength of Planet X is Wx/m.

C. The gravitational field strength of Planet X is 9.8 N/kg.

D. The gravitational field strength of Planet X is mWx.
Physics
1 answer:
Naya [18.7K]2 years ago
4 0

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

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Express the volume expansivity and the isothermal compressibility as functions of density rho and its partial derivatives. For w
kap26 [50]

Answer:

Explanation:

κ = 1/v  x dv/dp

= 1/ρ x (+δρ/ δp )_T

δρ = 1 , ρ = 100

44.18 x 10⁻⁶ = ( 1/100) x( 1 / δp)

δp = 10⁻² / (44.18 x 10⁻⁶ )

= .0226 x 10⁴

= 226 bar

7 0
2 years ago
When we draw a diagram of the forces acting on an extended object, the tail of the force vector for the weight should be at?
ser-zykov [4K]
Whenever we represent forces using vectors, we ensure that the vector begins at the point of force exertion. In the case of weight, the point of exertion of the force is known as the center of gravity, which is the point through which the weight of an object can be said to act. Therefore, the downward arrow representing the weight would begin at the center of gravity of the object.
6 0
2 years ago
A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.20 s. Then security
Brilliant_brown [7]

Answer: The ratio is 1.54

Explanation:

Firs, we have to find the relative speed of the man moving forward and backward.

Forward:

vf = vs + vm

vm = the man's speed

vs = the sidewalk's speed

vf = relative velocity moving forward

Because we don't know how much the man moved,

vf = distance (meters) / time (seconds)

vf = x / 2.20s

Backward:

vb = -vm + vs

vb = relative velocity moving backward

vf = distance (meters) / time (seconds)

vf = -x / 10.30s

We now divide the relative speeds

vf / vb = (x / 2.20) / (-x / 10.30)

We cancel the x

vf / vb = -10.3s / 2.2s = -4.68

vf = -4.68 . vb

We now substitute this in the equation we used for the forward travel

-4.68vb = vm + vs

Subtracting this from the backward travel equation

vb - (-4.68vb) = -vm - vm + vs -vs

5.68vb = -2vm

vb = -2vm / 5.68

Now, adding to the backward travel equation

vb + (-4.68vb) = -vm + vm + vs + vs

-3.68vb = 2vs

Using the two resulting equations

-3.68 . (-2 / 5.68) vm = 2vs

7.36 / 5.68 vm = 2vs

vm / vs = 1.54

7 0
2 years ago
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi
KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, A=7.85\times 10^{-5}\ m^2

Resistivity of copper, \rho=1.72\times 10^{-8}\ \Omega-m

The resistance of a wire is given by :

R=\rho\dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

8 0
2 years ago
A mine car, whose mass is 440kg, rolls at a speed of 0.50m/s on ahorizontal track, as the drawing shows. A 150kg chunk of coalha
ella [17]

Answer: 0.56 m/s

Explanation:

hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.

so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.

Now, the X component is definitely conserved here.

so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

440*0.5 + 150*0.8*cos(25) = 440*v_{final} + 150*v_{final} \\ => 220+120*cos(25) = (440+150)v_{final} => v_{final} = \frac{220+120*cos(25)}{590}  = 0.56 m/s

3 0
2 years ago
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