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Dmitry_Shevchenko [17]

2 years ago

15

Which force is represented by the arrow at A?

2 answers:

Nat2105 [25]2 years ago

7 0

Answer:

B) Force of gravity

Explanation:

As it is the force which acts downward on the body towards the ground, it is clearly the force gravity represented by the arrow at A.

Tom [10]2 years ago

4 0

Which force is represented by the arrow at A?

<em>B) Force of gravity</em>

<u>Let us understand the situation first and then we will understand which type of force is represented by arrow A </u>

<em>The</em><em> </em><em>line</em><em> </em><em>to</em><em> </em><em>which</em><em> </em><em>arrow</em><em> </em><em>A</em><em> </em><em>is</em><em> </em><em>being</em><em> </em><em>touched</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>ground</em><em> </em><em>and</em><em> </em><em>the</em><em> </em><em>block</em><em> </em><em>is</em><em> </em><em>being</em><em> </em><em>slid</em><em> </em><em>down</em><em> </em><em>a</em><em> </em><em>slope</em><em>.</em>
<em>The</em><em> </em><em>arrow</em><em> </em><em>A</em><em> </em><em>is</em><em> </em><em>representing</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>being</em><em> </em><em>applied</em><em> </em><em>by</em><em> </em><em>the</em><em> </em><em>earth</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>block</em><em> </em><em>and</em><em> </em><em>that</em><em> </em><em>force</em><em> </em><em>is</em><em> </em><em>Gravitational</em><em> </em><em>Force</em><em>.</em><em> </em>
<em>So</em><em>,</em><em> </em><em>Option</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>correct</em><em> </em><em>choice</em><em>!</em><em>!</em><em>~</em>

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A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago

The revolution of the earth around the sun demonstrate what motion?

olga2289 [7]

Answer:

Anticlockwise directions

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3 0

1 year ago

Friction is necessary when you are on a bike to stay

zzz [600]

Answer:

yes friction is needed hope this helps might of been to long tho

7 0

3 years ago

How much electrical energy is used by a 400 W toaster that is operating for 5

andrew11 [14]

Answer:

The answer is C. 120,000 J.

Explanation:

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2 years ago

the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m

olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

$x = \frac {-nmg}{k}$

Substituting into the formula, we have;

$x = \frac {-49*0.002*9.8}{24}$

$x = \frac {-0.9604}{24}$

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0

3 years ago