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3 years ago

The horizontal motion of a launched projectile affects its vertical motion.

Physics

1 answer:

Talja [164]3 years ago

8 0

The horizontal motion of a launched projectile affects its vertical motion is false

Answer: False

<u>Explanation: </u>

As the trajectory falls on a plane, all the components like displacement, velocity can be classified in two components termed as horizontal and vertical component.

In a projectile motion, the horizontal components of displacement and velocity remains constant and there is change only in the vertical component of displacement and velocity.

Both components are not dependent to each other. So, the horizontal motions of a launched projectile does not affect its vertical motion.

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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

4 0

3 years ago

Convert 141.2 kg to lbs and show all units and work

zavuch27 [327]

Your answer is 311.29271 lbs

5 0

3 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

6 0

3 years ago

What is characteristic of an opaque object.

stiks02 [169]

Answer:

allows no light to pass

Explanation:

5 0

2 years ago

Read 2 more answers

5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

8 0

3 years ago