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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

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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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

7 0

3 years ago

According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

$R=\dfrac{V}{I}$

The unit of voltage is volt and that of current is ampere

Unit of resistance :

$R=\dfrac{\text{volt}}{\text{ampere}}$

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0

3 years ago

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How does vacuum reduce thermal energy transfer

FrozenT [24]

The silver coating on the inner bottle prevents heat transfer by radiation, and the vacuum between its double wall prevents heat moving by convection. The thinness of the glass walls stops heat entering or leaving the flask by conduction.

3 0

3 years ago

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The two main states of mechanical energy are ___________ and potential energy.

shutvik [7]

<span>Potential energy and Kinetic energy</span>

6 0

3 years ago

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An object undergoes two successive displacements:

Gre4nikov [31]

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

$d_1=79 m$ at $16.9^{\circ}$

So its components are

$d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m$

2nd displacement is:

$d_2=16.7 m$ at $31.1^{\circ}$

So its components are

$d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m$

Therefore, the x- and y-components of the net displacement are:

$d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m$

Therefore, the magnitude of the final displacement is:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m$

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

8 0

3 years ago