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vladimir1956 [14]
3 years ago
7

The stationary object from which movement is

Physics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

I believe that the correct answer is D. reference point.

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If a man weighs 790 N on Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s2
sergey [27]

Answer:

The weight is   W_J  =  2087.9 \ N

Explanation:

From the question we are told that

   The weight of the man on earth  is  W =  790 \  N

  The acceleration due to gravity on Jupiter is a = 25.9 \ m/s^2

Gnerally the mass of the man is mathematically represented as

     m  =  \frac{W}{g}

=>   m  =  \frac{790 }9.8 }

=>   m  =  80.61 \  kg

Gnerally the weight in Jupiter is  

    W_J  =  m  *   a

=>  W_J  =  80.61  * 25.9

=>  W_J  =  2087.9 \ N

7 0
3 years ago
4 The kidneys, which remove dissolved wastes from the blood, are organs of the
eimsori [14]

Answer:

Urinary System

Explanation:

3 0
3 years ago
Read 2 more answers
An astronomer is trying to estimate the surface temperature of a star with a radius of 5.0×108m by modeling it as an ideal black
Jet001 [13]

Answer:

3.944 x 10⁴ K.

Explanation:

Let the surface temperature of star be T in absolute scale.

Total thermal radiation from its surface can be calculated with the help of

stefan's formula as follows

E = σ A T⁴

σ is a constant called stefan's constant and is equal to 5.67x 10⁻⁸ W m⁻²K⁻⁴

A is area of surface of star  and T is the surface  temperature.

Putting  the given values

E = 5.67 X 10⁻⁸ X 4 X 3.14 X ( 5 X 10⁸ )² X T⁴

= 1780.38  X 10⁸ T⁴

At a distance of 2.5 x 10¹³ m intensity will be calculated by dividing E with area of sphere having radius equal to distance , so

Intensity at given point

= E / 4πd²

= \frac{1780.38\times10^8T^4}{4\times3.14\times(2.5\times10^{13})^2}

= 22.68 T⁴ X 10⁻¹⁸

Putting the value of given intensity at that point

.055 = 22.68 x 10⁻¹⁸ T⁴

T⁴ = 24.25 X 10¹⁷

= 242.5 X 10¹⁶

T = 3.944 x 10⁴ K.

6 0
3 years ago
Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons and which contains O2 gas
Tems11 [23]
We are given with conditions on temperature, pressure and volume of gas. In this problem, we apply ideal gas law PV = nRT to solve what are asked.

1. PV = nRT ; n = m/MM
    16.5 x10^6 Pa * 0.003785 m3= n * 8.314 Pa m^3/ mol K * (23+273) K 
    n = 25.38 mol; mass = 812.08 grams. 
2. V at STP is 22.4 L/mol
3. PV = nRT = 1.5 atm *  0.003785 *1000 L = <span>25.38 mol * 0.0821 L atm/mol K * T ; T = 2.72 K
4. P * 55 m3/1000 = 25.38 mol * </span><span>8.314 Pa m^3/ mol K</span><span> * (24+273) K
  P = 1139. 45 kPa</span>
5 0
3 years ago
on a aircraft carrier a jet is slowed from 105 mph to a stop in 2 seconds. What is the rate of deceleration?
Bess [88]
105/2 = 52.5/1

The rate of deceleration is 52.5 mps (miles per second).
6 0
3 years ago
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