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3 years ago

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1) Which is a characteristic of a metal? A) It looks dull. B) It feels brittle. C) It is a good insulator. D) It is a good condu

ctor of electricity. 2)

1 answer:

max2010maxim [7]3 years ago

6 0

D) It is a good conductor of electricity

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An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider compl

Novosadov [1.4K]

Answer:

T = 2.82 seconds.

The frequency $\mathbf{f = 0.36 \ Hz}$

Amplitude A = 25.5 cm

The maximum speed of the glider is $\mathbf{v = 56.87 \ rad/s}$

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

$T = \frac{31}{11}$

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

$f = \frac{1}{T}$

$f = \frac{1}{2.82}$

$\mathbf{f = 0.36 \ Hz}$

The amplitude is determined by using the formula:

$A = \frac{d}{2}$

The limits that the spring makes the oscillations are from 10 cm to 61 cm.

The distance of the glider is, d = (61 - 10 )cm = 51 cm

Replacing 51 for d in the above equation

$A = \frac{51}{2}$

A = 25.5 cm

The maximum speed of the glider is:

$v = A \omega$

where ;

$\omega = \frac{2 \pi}{T}$

$\omega = \frac{2 \pi}{2.82}$

$\omega = 2.23 \ rad/s$

$v = A \omega$

$v = 25.5 *2.23$

$\mathbf{v = 56.87 \ rad/s}$

3 0

3 years ago

The student moves his leg 34 cm sideways as shown.

Nana76 [90]

(i) • there is force applied to an objects

• the object moves

• the object moves in the same direction as the direction of the force

(ii) workdone = force x distance

= 23 x 34

= 782Joules

8 0

3 years ago

Read 2 more answers

Explain why an object’s weight is dependent upon where in the universe it is located.

strojnjashka [21]

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the law of universal gravity

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]$

Now replacing the values in the equation we have:

$F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]$

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

$g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]$

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a radius of 71492 [km]. Find the force exerted by the planet upon the person.

$F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} } \\F= 1734.73[N]$

We will calculate the value of the gravity accelaration of Jupiter. Let's check

$g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\$

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

6 0

3 years ago

La masa de un objeto en la tierra es de 100 kg.

andreev551 [17]

Answer:

a) 980 N

b) 100 kg

c) 163.33 N

Explanation:

a)

The weight of the object on Earth is the product of its mass times the acceleration due to gravity, that is:

$W=m\,g=(100\,\,kg)\,g=(100\,\,kg)\,(9.8\,\,m/s^2)=980\,\,N$

b)

The mass of the object on the moon, is exactly the same (100 kg) since mass is an inherited value of the object.

c)

Considering that the acceleration of gravity on the Moon is about 1/6 that we have on earth (g), then the weight of the object on the Moon becomes:

$W=m\,\frac{g}{6} =\frac{980}{6}\,\,N=163.33\,\,N$

3 0

2 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

2 years ago