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san4es73 [151]
2 years ago
15

To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo

u choose to throw a rubber ball or a beanbag of equal size and weight?
Physics
1 answer:
Setler [38]2 years ago
8 0

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\  P = -2mv_{i}

For Beanbag

value of impulse will large if velocity is zero.

v_{f}=0\\ P = -mv_{i}

Explanation:

You might be interested in
A 2.29N force is applied to the plunger of a hypodermicneedle.
Ede4ka [16]

Answer:

0.04651 N

Explanation:

F_1 = Force on plunger = 2.29 N

d_1 = Diameter of plunger = 1.27 cm

d_2 = Diameter of needle = 0.181 cm

F_2  = Force on needle

From Pascal's law we have

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1\times A_2}{A_1}\\\Rightarrow F_2=\dfrac{2.29\times \pi\dfrac{0.181^2}{4}}{\dfrac{\pi 1.27^2}{4}}\\\Rightarrow F_2=0.04651\ N

The force with which the fluid leaves the needle is 0.04651 N

6 0
3 years ago
The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door,
ValentinkaMS [17]

Answer:

T0=390 degF

Explanation:

The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door, an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.

If T is temperature, then by Newton’s law

dTdt=-k(T−T0)

Where

T0− is temperature of oven  

∫\frac{dT}{T-T0} =\int\limits^0.5_0 {-k} \, dt

ln(T-T0), from 110 to 70=-kt, from 1/2 to 0

ln(110-T0)-ln(70-T0)=-k(1/2-0)

2in\frac{110-T0}{70-T0} =-k\\k=-2ln\frac{T0-110}{T0-70} ...........................1

integrating the LHS of the equation from 110 to 145F

∫∫∫\frac{dT}{T-T0} =\int\limits^1_0.5 {-k} \, dt

ln(145-T0)-ln(110-T0)=-k(1-1/2)

2ln145-T0/(110-T0)=-k

k=-2lnT0-145/(T0-110).......................2

couplijng equatiuon 2 with 1

(T0-110)^2=(T0-70)(T0-145)

T0^2-220T0+12100=T0^2-215T0+10150

5T0=1950

T0=390 degF

5 0
3 years ago
Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea
notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, t=0.50\ s

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, v=0\ m/s

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, a=g=-9.8\ m/s^2

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

v=u+at

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0
3 years ago
Read 2 more answers
A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
3 years ago
A man jogs at a speed of 1.6 m/s. His dog
FromTheMoon [43]
I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m
5 0
3 years ago
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