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monitta
2 years ago
5

A man is holding a 5kg weight ball. What is the weight of the ball?

Physics
1 answer:
jonny [76]2 years ago
3 0

Answer:

1000kg

Explanation:

I'm smart I am a genius I know everything so I know it is 1000kg

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
The nearly circular path that earth takes around the sun is called
Sati [7]
It is called an "orbit". Repeating and regular path that one object in space takes around in another one. This object ca be satellite or natural satellite like our moon. Planets, comets, asteroids and other objects in the solar system orbit the sun. Orbits come in different forms or shapes. All orbit are elliptical which means they are ellipse, similar to oval.
4 0
3 years ago
What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0
2 years ago
HlEELPPI NEED ANSWER 
natka813 [3]
THE ANSWER IS C ;) !





















3 0
4 years ago
Read 2 more answers
Initially at rest a single-stage rocket is launched vertically from the ground. The rocket’s thrust overcomes gravity and provid
Lelechka [254]

Answer:

Explanation:

Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds

s = 1/2at²

= .5 x a x 100

=50a

Velocity after 10 s

u = at = 10a

Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.

s = ut + 1/2 gt²

50a = -10ax20 + .5 x 10x400

250a = 2000

a = 8 m s⁻² .

Velocity after 10 s

= at = 80 m/s

further height reached with this speed under free fall

h = v² / 2g

= 80 x 80 / 2 x 10

= 320 m

height achieved under acceleration

= 50a

= 50 x 8 = 400m

Total height

= 320 + 400= 720 m

velocity after falling from 720 m

v² = 2gh = 2 x10 x 720

v = 120 m/s

7 0
3 years ago
Read 2 more answers
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