What is the chemical name for lime

A sphere moves in simple harmonic motion with a frequency of 4.40 Hz and an amplitude of 3.50 cm. (a) Through what total distanc

Anna [14]

Answer:

It will move a distance of 4 * 3.5 = 4 cm in one cycle.

zero to max A, back to zero, zero to min A, back to zero (4 * A)

6 0

2 years ago

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

3 years ago

A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,

fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

$p=mv$

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

$m'=2m$

therefore, the new momentum is

$p'=m'v=(2m)v=2(mv)=2p$

so, the momentum has also doubled.

7 0

3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago

Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx

Luda [366]

Answer:

your question in not complete.

you need to the high too.

7 0

3 years ago