Ask question

Ask question

All categories

2 years ago

6

An iron nail is driven into a block of ice by a single blow of a hammer. The hammerhead has a mass of 0.5 kg and an initial spee

d of 3.2 m/s. Nail and hammer are at rest after the blow. How much ice melts? Assume the temperature of both the ice and the nail is 0◦C before and after. The heat of fusion of ice is 80 cal/g. Answer in units of g.

1 answer:

larisa [96]2 years ago

5 0

Answer:

The ice melts mass is:

$m_g=7.6x10^{-3} g$

Explanation:

Kinetic Energy

$K_E = 1/2*m*v^2$

$K_E = 1/2*0.5kg*(3.2m/s)^2$

$K_E=2.56 kg*m^2/s^2$

Heat gained by ice= mass(g) x 80 cal

( 1 cal = 4.184 *10^7er or g cm^2/ sec^2)

Assuming no loss in heat, in the motion so both continue with temperature 0~C

To find so the mass (gm) of ice melted

$m_g= 1/2 *(0.5*1000)*(3.2m/s)^2* 100*100 / (80*4.184*10^7)$

$m_g=7.6x10^{-3} g$

You might be interested in

A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal

daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0

2 years ago

PLS ANSWER FAST WILL GIVE BRAINLY TIMED TEST

solong [7]

A=F/m
a=(3000000)/(20000)
a=15 m/s^2

4 0

2 years ago

Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$

6 0

2 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

2 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

likoan [24]

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it $W_{a-b}=-1.9*10^{-8}J$

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

$W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}$

Now substitute the given values

So

$U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J$

3 0

3 years ago