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Natasha2012 [34]

3 years ago

13

Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image

below in newtons. What would be the force exerted by gravity on 9.0 kg?
a. 9.0n
b.9.8n
c. 81n
d.88 n

1 answer:

nekit [7.7K]3 years ago

5 0

d.88 n

Explanation:

The force exerted by gravity on the mass is 88N;

Gravitational force is usually experienced by two bodies with masses.

This force causes all objects on earth to accelerate at 9.8m/s².

Force of gravity = mass x acceleration due to gravity

In this case, the force exerted on the object by gravity will be the weight of the object.

Weight is a force applied gravity on a body:

Weight of the masses = Force of gravity = mass x acceleration

Given mass = 9kg

Force of gravity = 9 x 9.8 = 88.2N approximately 88N

learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

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The resistance of a conductor is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the resistivity of the material
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A is its cross-sectional area

We can use this formula to solve both parts of the problem.

a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is
$r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m$
And its cross-sectional area is
$A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2$
The copper resistivity is $\rho=1.68 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}= 8.57 \cdot 10^{-2} \Omega$

b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
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The iron resistivity is $\rho = 9.71 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
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3 years ago

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Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

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Hello!

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