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3 years ago

How can you double the acceleration of an object if you cannot alter the object’s mass?

Physics

2 answers:

Lyrx [107]3 years ago

7 0

Answer:

If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.

Explanation:

serious [3.7K]3 years ago

4 0

you can double the net force

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It wouldn't indicate the same source if the paint was from a different place. If the paint was the same brand and color it would be but if not then no

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3 years ago

4- Protein foods?? pls answer these i really need these

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3 years ago

What is occurring when a light wave goes through a pane of glass in a window?

ycow [4]

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Explanation:

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2 years ago

a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su

Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

$U=(M+m)gh$ (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

$h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m$

The potential energy is:

$U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J$

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

$U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}$

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

$Mv_1+mv_2=(M+m)v$ (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

$M(0)+mv_2=(M+m)v$

$v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}$

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

$I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}$

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

$T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N$

The force on the cord after the impact is 2.59N

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3 years ago

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puteri [66]

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3 years ago