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2 years ago

7

Calculate the weight of a 1 kg mass at earth's surface. The mass of the of the Earth's surface if the mass of the earth is 6 x 1

024 kg and its radius is 6.4 x 10 m.

1 answer:

yuradex [85]2 years ago

4 0

Answer:

9.8N

Explanation:

Here we can get gravitational acceleration according to the place where object is placed by bellow equation

g = GM/R²

g - Gravitational Acceleration

G - Gravitational constant (6.67×10-11)

R - Distance ( Radius )

g = 6.67 × 10-11 × 1024 /(6.37×106)²

g = 9.8 m/s²

There for

Weight = Mass × Gravitational acceleration

= 1×9.8

= 9.8 N

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A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$

6 0

2 years ago

3. An object of mass 90 kg travels down a slide.

Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

$E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]$

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

$E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]$

4 0

2 years ago

Likurg_2 [28]

When you know both the speed and direction of an object’s motion, you know the velocity of an object. speed in a given direction is called velocity
hope this helps :)

8 0

2 years ago

True or False: The energy increase of an object acted on only by a gravitational force is equal to the product of the object's w

tamaranim1 [39]

Answer:

False.

Explanation:

The statement shown in the question above is false and this can be confirmed by Newton's law on universal gravitation. According to Newton, the gravitational force exerted on any body is proportional to its weight, but the distance that the object travels when falling is disproportionate. In addition, if the force resulting from the weight of the object and its displacement has an angle of 0º, the weight force of that object will provide an increase in kinetic energy.

4 0

3 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

2 years ago