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alina1380 [7]
3 years ago
13

An object has a mass of 2,000kg. what is its weight on earth? show your work​

Physics
1 answer:
Rudiy273 years ago
8 0

Answer:

Answer: 19,600 N

Explanation:

Weight is a force, and force equals mass times acceleration. So the formula for weight is w = m • g

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
Is Radioactive decay a way to find the relative age of a rock?<br><br> True<br> False
galina1969 [7]
To establish the age of a rock or a fossil, researchers use some type of clock to determine the date it was formed. Geologists commonly use radiometric dating methods, based on the natural radioactive decay of certain elements such as potassium and carbon, as reliable clocks to date ancient events.
5 0
3 years ago
Teachers are interested in knowing what study techniques their students are utilizing. The researchers randomly select every 10t
uranmaximum [27]

Answer:

Simple Random Sample (SRS)

Explanation:

5 0
3 years ago
Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

  • V = Linear speed
  • W = Angular speed
  • r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2\pi/T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x 10^{-4} Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x  10^{-4} x 149.6 x 10^{6}

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4\pi ^{2}r^{3}) / GT^{2}

M = (4 x 9.8696 x 3.35 x 10^{24}) / (6.67 x 10^{-11} x 7.68 x 10^{11}<em>)</em>

<em>M = 1.32 x </em>10^{26} / 51.226

M = 2.58 x 10^{24} Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

#SPJ1

3 0
2 years ago
If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose
Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0
3 years ago
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