Ask question

Ask question

All categories

2 years ago

5

A 5.00-g lead BB moving at 44.0 m/s penetrates a wood block and comes to rest inside the block. If half of its kinetic energy is

absorbed by the BB, what is the change in the temperature of the BB? The specific heat of lead is 128 J/kg ∙ K

1 answer:

guapka [62]2 years ago

3 0

Answer:

ΔT=3.781 Kelvin

Explanation:

Given data

m=5.00g 0r 0.005kg

v=44.0 m/s

c (specific heat)=128 j/kg

Kinetic Energy= $\frac{1}{2}mv^{2}$

KE= $\frac{1}{2}(0.005)(44)^{2}$

KE=4.84 joules

1/2 Kinetic energy goes to heat=1/2×4.84

=2.42 J

Q=mcΔT

ΔT= $\frac{Q}{mc}$

ΔT= $\frac{2.42}{0.005*128}$

ΔT=3.781 K

You might be interested in

main idea which will change if you turn up a radios volume: wave velocity, intensity, pitch, frequency, wavelength, loudness.

Nadusha1986 [10]

Loudness is one of answers

7 0

2 years ago

The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac

Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

$Pressure = h\cdot\rho\cdot g$

Case 1 - Mercury in the tube

$Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1$

Case 2 - Water in the tube

$Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2$

Since atmospheric pressure is same

. $P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g$

or, $h_2=\frac{h_1\rho_1}{\rho_2}$

$Given\ h_1= 0.76\ m,\rho_1=13.6\cdot\rho_2$

∴ $h_2=0.76\cdot13.6=10.336\ m$

Hence height of the water column =10.336 m

6 0

2 years ago

Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?

Svetradugi [14.3K]

Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>

6 0

2 years ago

Read 2 more answers

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is friction work if any

$490=360+W_{f}$

$W_{f}=130 J$

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0

2 years ago

Which equation could be rearranged to calculate the frequency of a wave?

777dan777 [17]

Answer:

wavelength = speed/frequency

Explanation:

Required

Determine which of the options can be used to calculate frequency

The relationship between wavelength, speed and frequency is as follows;

$Frequency = \frac{Wave\ Speed}{Wave\ Length}$ ---- Equation 1

When option (1), (2) and (4) are rearranged, they do not result in the above formula; only option (3) does

Checking option (3)

$Wave\ Length = \frac{Speed}{Frequency}$

Multiply both sides by Frequency

$Wave\ Length * Frequency = \frac{Speed}{Frequency} * Frequency$

$Wave\ Length * Frequency = Speed$

Divide both sides by Wave Length

$\frac{Wave\ Length * Frequency}{Wave\ Length} = \frac{Speed}{Wave\ Length }$

$Frequency = \frac{Speed}{Wave\ Length }$ --- Equation 2

<em>Comparing equation 1 and 2; both equations are the same.</em>

<em>Hence, option (3) answers the question</em>

7 0

2 years ago