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malfutka [58]
3 years ago
6

A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling

Physics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

Solving for time :

(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known

2. The unknown constant that we want to solve)

s = (1/2)(u+v)t <--- one of the formulas

from linear motion

s (distance) = 0.05m

u (initial velocity) = 100m/s

v (final velocity) = 0 m/s (it stops)

t (time taken for change in velocity) = to be found

0.05 = (1/2)(100+0)t

t = 0.001 seconds

Solving for the resistant force :

Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.

When the bullet stops :

F net = 0

F r = F imp

F r = (mu -mv)/t

F r = (0.01x100-0.01x0)/0.001

F r = 1/0.001

F r = 1000N

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A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a
anastassius [24]

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

5 0
2 years ago
Swimmers at the beach are tanning on towels. Which method of heat transfer is responsible for their tan? *
Free_Kalibri [48]

Answer:

ratiation

Explanation:

3 0
2 years ago
Read 2 more answers
URGENT
Advocard [28]

Answer:

give \\ mass(m) = 1350kg  \\ acceleration(a) = 1ms {}^{ - 2}  \ \\ sln\ \\  from \: our \: formula \\  \: f = ma \\1350kg \times 1ms {}^{ - 2}  \\ f = 1350newton

the force you applied to your car =1350N

5 0
3 years ago
A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
The wind increases to 14 mph from the north. Now what is your airspeed and what direction are you flying? If your destination is
ladessa [460]

Answer:

I-

Explanation:

...

7 0
3 years ago
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