Calculate the kinetic energy of a 20-kg sled moving 28.0 m/s. Show your work in the space to the right.

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

A portable CD player uses two 1.5 volt batteries. if the current in the CD player is 2amps ,what is resistance?

chubhunter [2.5K]

resistance=voltage/current

1.5/2=0.75ohms

7 0

3 years ago

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

The mass of the metal is $M = 60 \ kg$

The specific heat of the metal is $c_p = 0.1027 kcal/(kg \cdot ^oC)$

The mass of the oil is $M_o = 810 \ kg$

The temperature of the oil is $T_o = 35^oC$

The specific heat of oil is $c_o = 0.7167 kcal/(kg \cdot ^oC )$

The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

Heat lost by metal = heat gained by the oil

So

The quantity of heat lost by the metal is mathematically represented as

$Q = - Mc_p \Delta T$

=> $Q = -Mc_p (T_m - T_c)$

Where $T_ m$ the temperature of metal before immersion

The negative sign show heat lost

The quantity of gained t by the metal is mathematically represented as

$Q = M_o c_o \Delta T$

=> $Q = M_o c_o (T_c - T_o)$

So

$Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

$- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=> $T_m = 376.8 ^o C$

6 0

3 years ago

In making the determination that bullets, shells, or cartridge cases were fired from a specific weapon, the criminalist would us

ss7ja [257]

Answer:

Comparison Microscope

Explanation:

The Comparison Microscope allows for comparison between two objects or samples by placing them side by side.

It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

7 0

3 years ago