1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
suter [353]
3 years ago
9

What would happen if you threw a grenade into a woodchipper?

Physics
2 answers:
Olin [163]3 years ago
7 0

Answer:it explodes and splinters everywhere

Explanation:

sammy [17]3 years ago
3 0
I'm not sure but it might explode and make the wood catch fire
You might be interested in
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
If 100 J of electrical energy enter the bulb and 5 J of light energy leave the bulb, how many joules of heat energy leave the bu
Novay_Z [31]

As per energy conservation we know that

Energy enter into the bulb = Light energy + Thermal energy

so now we have

energy enter into the bulb = 100 J

Light energy = 5 J

now from above equation we have

100 = 5 + heat

Heat = (100 - 5) J

Heat = 95 J


6 0
3 years ago
Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If
jeka57 [31]

Let F_{1}=F_{2}=F.

Normal force equals (using Newton's third law) N=mg+F\sin30^{o}-F\sin37^{o}.

F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o}), but F_{f}\leq F(\cos30^o+\cos37^o) for all F_{f} (in order to start moving the break). Therefore F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o}), solving for F: F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}

7 0
2 years ago
What is the center of gravity
Gala2k [10]
The center of gravity is a SIKKE bro you really thought I’d give the answer of something so simple
7 0
3 years ago
Read 2 more answers
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
2 years ago
Other questions:
  • What properties of sound determine the volume of sound?
    10·1 answer
  • How did you know if an electric current is<br> flowing in a lightbulb?
    10·1 answer
  • A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.
    7·1 answer
  • To calculate the heat needed to melt a block of ice at its melting point what do you need to know
    8·1 answer
  • What is momentum in physics?
    15·2 answers
  • Explain how heat transfer is occurring through convection, radiation AND conduction in this picture.
    8·1 answer
  • Driving to work one morning, you get a flat tire. When using the car jack, you apply 120N of force to the jack and the jack in t
    6·1 answer
  • In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the cir
    11·1 answer
  • Which statement best describes energy and matter in a closed system?
    9·1 answer
  • A stone is thrown at 45 degree with velocity 10m/sec.calculate the range of projectile​
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!