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4 years ago

Can you change the weight of something without changing the mass?

Physics

1 answer:

svlad2 [7]4 years ago

3 0

yes because your mass doesn't change but your weight can

ex- if you travel to Saturn your weight would change but your mass would stay the same

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A boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radiu

irina [24]

Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

ω² = 926

ω = 30.4 rad /s

angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

time taken = θ / ω

= 126.6 / 30.4

= 4.16 s .

3 0

3 years ago

A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

6 0

3 years ago

Is the process of introducing a non blood fluid into the blood

vladimir1956 [14]

The lymphatic system

4 0

4 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago

Please subscribe wonder melodies please on you tube channel

Anuta_ua [19.1K]

Sure , I'm subscribing!!!

8 0

3 years ago

Read 2 more answers