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1 year ago

5

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block c

an move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

1 answer:

schepotkina [342]1 year ago

8 0

After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

Here we will apply the principle of impulse and momentum.

Horizontal component:

-mv₀ cos30⁰ + RxΔt =0

RxΔt = mv₀sin30⁰

= 0.001941 × 1400sin30⁰

RxΔt = 1.3587 lb.s

Vertical component:

-mv₀cos30⁰ + RyΔt = -mv'

RyΔt = m( v₀cos30⁰-v')

RyΔt = 0.001941(1400cos30⁰ - 14.968)

= 2.32 lb.s

Learn more about impact here:

brainly.com/question/15008937

#SPJ4

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Why is it that the weight of an object weighing 1N air, weighs more when immersed in water ?

Anni [7]

There is no "why", because that's not what happens. The truth is
exactly the opposite.

Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.

The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air. So the force that lifts the
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appears to weigh less in the water.

4 0

3 years ago

aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0

2 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0

2 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

3 years ago

A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of

vaieri [72.5K]

Answer:

(a) P = 459.055 N.

(b) the refrigerator tips.

Explanation:

Given, the angle of ramp is 20°.

When the weight of refrigerator is resolved in directions parallel and perpendicular to ramp, 75×g×sin(20°) and 75×g×cos(20°).

⇒ normal contact force is 75×g×cos(20°).

⇒ frictional force is 0.3×75×g×cos(20°) = 207.414 N

so, total opposite force is 207.414 + 75×g×sin(20°) = 459.055 N.

so, the force needed is P = 459.055 N

And as the moment due to both opposite force and P force are in same direction the refrigerator tips rather than just sliding.

4 0

3 years ago