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Nady [450]
3 years ago
13

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 965m/s. the mass of the hunter (in

cluding his gun) is 72.5kg, and the hunter holds tight to the gun after firing it. find the recoil velocity if he fires the velocity horizontally. enter as positive. find the recoil velocity if he first the rifle at 56.0° above the horizontal. enter as positive.
Physics
1 answer:
weqwewe [10]3 years ago
3 0

When the gun is fired horizontally :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 m/s

M = mass of the hunter including gun  = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (965) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.056 m/s

so recoil velocity comes out to be 0.056 m/s



When the gun is fired at angle 56.0⁰ above the horizontal :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 Cos56 = 539.62 m/s

M = mass of the hunter including gun  = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (539.62) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.031 m/s

so recoil velocity comes out to be 0.031 m/s




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Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

Speed  = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

  Speed  = \frac{300}{18}  = 16.67m/s

3 0
3 years ago
You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?
rodikova [14]

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

3 0
3 years ago
Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
B. The silica cylinder of a radiant wall heater is 0.6 m long
SIZIF [17.4K]

So,  If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

  • ε = emissivity = 1 (since we are not given),
  • σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
  • A = surface area of cylindrical wall heater = 2πrh where
  • r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
  • h = length of heater = 0.6 m, and
  • T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π  × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72  × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

6 0
2 years ago
URGENTEEEEEEEEEEEEEE
OverLord2011 [107]

Answer:

no habl

Explanation:

6 0
3 years ago
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