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Juliette [100K]
3 years ago
11

Two spaceships that have different masses but rocket engines of identical force are at rest in space. if they fire their rockets

at the same time, which ship will speed up faster
Physics
1 answer:
NISA [10]3 years ago
5 0
Force is calculated F=m×a.
If both ships speed up with the same force, but have a different mass, This means that a also has to be different. If F is the same but ship a has a bigger mass(m) than ship b, then the acceleration(a) of ship b has to be bigger so F of each ship is the same. So the ship with the smaller mass will speed up faster.
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How much time does it take for tweety’s bird cage to hit the ground after it was dropped if it reached a velocity of 22 meters p
guajiro [1.7K]
It matters on the weight

7 0
3 years ago
100 J OF HEAT IS PRODUCED EACH SECOND IN A 4 COULUMB RESISTER. THE POTENTIAL DIFFERANCE ACROSS THE RESISTER WILL BE
igomit [66]

Answer:

The correct answer is "20 Volts".

Explanation:

Given:

Heat,

H = 100 J

Resistance,

R = 4 Ω

As we know,

⇒ P=\frac{v^2}{R}

By putting the values, we get

⇒ 100=\frac{v^2}{4}

⇒  v^2=100\times 4

⇒       =400

⇒    v=\sqrt{400}

⇒       =20 \ volts

6 0
3 years ago
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
statuscvo [17]

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required = \frac{(14.644 x 10^6)}{784}

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

3 0
3 years ago
Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen
Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0
2 years ago
Can someone help :3 dis is hard
Murrr4er [49]
C is the correct answer beacause it shows where it is happening in this cas “here”.
8 0
3 years ago
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