Two cars collide head on while each is traveling at 50 km/h . Suppose all their kinetic energy is transformed into the thermal energy of the wrecks. What is the temperature increase of each car? Part A You can assume that each car's specific heat is that of iron

Answer:

Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.

The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.

Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.

A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.

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Answer:

The correct answer is - 93.24×10^4 kg/m^3.

Explanation:

Given:

height of cylinder: 50.5 mm

diameter = 52.0

then radius will be diameter/2 = 52/2 = 26

Formula:

Density = mass/ volume

Volume = πr^2h

solution:

Now the volume of a cylinder is v = (22/7)×r^2×h

= 22/7×26×26×50.5

= 107261.59 mm^3  

Now volume in cubic meter V =10.7261 ×10^(-5) m^3

So density d = m/V = 1/(10.7261 ×10^(-5))

Or d = 93.24×10^4 kg/m^3

Answer:

[tex]21.6\ \text{kg m}^2[/tex]

[tex]3.672\ \text{Nm}[/tex]

[tex]54.66\ \text{revolutions}[/tex]

Explanation:

[tex]\tau[/tex] = Torque = 36.5 Nm

[tex]\omega_i[/tex] = Initial angular velocity = 0

[tex]\omega_f[/tex] = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

[tex]\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2[/tex]

Torque is given by

[tex]\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2[/tex]

The wheel's moment of inertia is [tex]21.6\ \text{kg m}^2[/tex]

t = 60.6 s

[tex]\omega_i[/tex] = 10.3 rad/s

[tex]\omega_f[/tex] = 0

[tex]\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2[/tex]

Frictional torque is given by

[tex]\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}[/tex]

The magnitude of the torque caused by friction is [tex]3.672\ \text{Nm}[/tex]

Speeding up

[tex]\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}[/tex]

Slowing down

[tex]\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}[/tex]

Total number of revolutions

[tex]\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}[/tex]

[tex]\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}[/tex]

The total number of revolutions the wheel goes through is [tex]54.66\ \text{revolutions}[/tex].

Answer:

Universe

Explanation:

I took the quiz.

Answer:

72.57° North of east

Explanation:

From the given information:

We can compute the velocity plane that is related to the ground in air in the North direction as;

[tex]v^{\to} _{PG} = v \\ \\ v^{\to} _{PG,x} = 0 \\ \\ v^{\to} _{PG,y} = v[/tex]

However, the velocity of the wind-related to the ground from the NorthEast  direction is;

[tex]v^{\to}_{wG}=100 \ km/h \\ \\ \text{from North East} \\ \\ v_{wG,x} = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_{wG,y} = (-100 \ km/h ) sin 45 = -70.7 km/h[/tex]

Now,

Since the plane is moving with a 236 km/h speed in the Northeast direction;

Then;

[tex]v^{\to} _{pw} = 236 \ km/h \\ \\ v^{\to} _{pw.x} = (236 m/s) cos \theta \\ \\ v^{\to} _{pw,y} = (236\ m/s) sin \theta \\ \\ v_{pG,x} = v_{pw,x} + v_{w G,x} \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = \dfrac{70.7 \ km/h}{236 \ km/h} \\ \\ \theta = cos^{-1} (0.2996) \\ \\ \mathbf{\theta = 72.57}[/tex]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

Distance travelled in one round will be equal to the circumference of the disc i.e [tex] 2 \pi r[/tex]

Radius=50cm

Circumference= [tex] 2 \times \frac{22}{7} \times 50cm=> \frac{2200}{7} cm[/tex]

If the disc rotates at speed of 100rpm that means it completes 100 rotation in a minute(60 second)

So, in 30 seconds it will complete 50 rotation.

1 rotation = [tex] \frac{2200}{7} cm [/tex]

[tex] 50 rotation=\frac{2200}{7} \times 50cm \\\\ \frac{110000}{7} cm[/tex]

Break it in decimal.

Your answer will be 15714.2 cm

Answer:

h=0.142m=14.2cm

Explanation:

ρ=8.96 g/cm3=8960 kg/m3

d=0.1m

ρ=m/V - - >ρ=m/[π*((d/2)^2) *h] - - >

8960=10/[π*((0.1)^2) *h] - - >

h=0.142m

Answer:

Current is calculated as the amount of charge that pass  a point on a circuit per time.

Explanation:

Electric current is defined as the amount of charge that passes a point on a circuit per unit time. The mathematical definition of electric current is given by :

[tex]I=\dfrac{q}{t}[/tex]

Where

q is a charge (Coulomb)

t is time (in seconds)

The SI unit of electric current is A. It is equivalent to C/s. So, the correct option is (A).

Answer:

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

Answer: Mercury

Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.

Answer: 51

Explanation:

Given

Output is 23 V

The square armature side is [tex]a=5.1\ cm[/tex]

Magnetic field [tex]B=0.5\T[/tex]

Rate of revolution [tex]n=55\ rev/s[/tex]

Angular speed

[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]

Peak voltage is given by

[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]

So, there are approximate 51 loops

Answer:

  a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]  ,       t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Explanation:

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

         Wₓ - T = m a

Y axis  

         N- W_y = 0

         N = W_y

for pulley

          ∑τ = I α

           T R = (½ m_p R²) α

let's use trigonometry for the weight components

         sin β = Wₓ / W

         cos β = W_y / W

         Wx = W sin β

angular and linear variables are related

          a = α R

          α = a / R

we substitute and group our equations

         W sin β - T = m a

         T R = ½ m_p R² (a / R)

         W sin β - T = m a

                        T = ½ m_p a

we solve the system of equations

         W sin β = (m + ½ m_p) a

          a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]

let's find the time to travel the distance (d) through the block

          x = v₀ t + ½ a t²

          d = 0 + ½ a t²

          t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Answer:

FORCE as for my answer....

Answer:

okay

Explanation:

please I don't know

This question is incomplete, the complete question is;

A rifle fires a 2.10 × 10⁻² kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 9.10 × 10⁻² m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answer:

the spring constant is 303.5 N/m

Explanation:

Given the data in the question;

There is a potential energy associated with the spring;

we know that potential energy = kinetic energy

mgh = [tex]\frac{1}{2}[/tex]kx²

where k is the spring constant and x is the compression

as the pallet is fired, the spring gives kinetic energy which is converted into gravitational potential energy

so

m = 2.10 × 10⁻²

g = 9.81 m/s²

h = 6.10 m

x = 9.10 × 10⁻² m

we substitute

mgh = [tex]\frac{1}{2}[/tex]kx²

2.10 × 10⁻² × 9.81 × 6.10 =  [tex]\frac{1}{2}[/tex] × K × ( 9.10 × 10⁻² )²

1.256661 = 0.0041405 × K

K = 1.256661 / 0.0041405

K = 303.5 N/m

Therefore, the spring constant is 303.5 N/m

Answer:

D) Report the problem to your supervisor.

Explanation:

This is probably the most efficient way get them to stop or to get them to follow the rules <3

Answer:

BI/BII = 1

Explanation:

The magnetic field due to a solenoid is given by the following formula:

[tex]B = \mu nI\\[/tex]

where,

B = Magnetic Field due to solenoid

μ = permeability of free space

n = No. of turns per unit length

I = current passing through the solenoid

Now for the first solenoid:

[tex]B_1 = \mu n_1I_1 \\[/tex]

For the second solenoid:

[tex]B_2 = \mu n_2I_2\\[/tex]

Dividing both equations:

[tex]\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\[/tex]

here, no. of turns and the current passing through each solenoid is same:

n₁ = n₂ and I₁ = I₂

Therefore,

[tex]\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\[/tex]

BI/BII = 1

Answer:

D. Electrons are tightly bound to the nuclei.

Explanation:

In an insulator, the electrons of the outer most shell are bound with a very high electrostatic forces coming from the nucleus of each atom so electrons cannot flow around all atoms making up the material as in a conductor.

The characteristic of a good insulator is Electrons are tightly bound to the nuclei. (option d)

In a good insulator, electrons are tightly bound to the nuclei of their atoms. This means that they are not free to move around within the material, unlike conductors where electrons are relatively loosely bound and can move freely. Due to this strong binding, electrons in insulating materials cannot carry an electric charge or energy easily from one atom to another.

When an electric field is applied to an insulator, the electrons may experience a small displacement within their respective atoms, but they generally do not move from one atom to another or flow through the material like they would in a conductor. As a result, insulators prevent the flow of electric current and are used to isolate or protect conductive elements from accidental contact.

So, the correct answer is D. Electrons are tightly bound to the nuclei.

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Answer:

D) Mechanical waves require a medium for transmission (wire, air, etc.) as opposed to electromagnetic which travel through empty space - light, radio, etc.)

Answer:

   a = 7.29 m / s²,      T = 0.40 N

Explanation:

To solve this exercise we must apply Newton's second law to each body

The needle

              W -T = m a

              mg - T = ma

The spool, which we will approach by a cylinder

             Σ τ = I α

             T R = I α

the moment of inertia of a cylinder with an axis through its center is

             I = ½ M R²

angular and linear variables are related

            a = α R

            α = a / R

we substitute

           T R = (½ M R²) a / R

            T = ½ M a

we write our system of equations together

              mg - T = m a

                      T = ½ M a

we solve

              m g = (m + ½ M) a

              a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]

let's calculate

              a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]

              a = 7.29 m / s²

now we can look for the tension

              T = ½ M a

              T = ½ 0.110 7.29

               T = 0.40 N

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

Answer:

iEvaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)

Explanation:

Answer:

yes

it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

Answer:

hi I thinks its number 3

Explanation:

hope you have a nice day

Answer:

Explanation:

I jus did it on usatestprep

Answer:

D

Explanation:

From the information given:

The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]

Disk radius (r) = 0.2 m

The block Initial velocity is:

[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]

Change in the block's angular speed is:

[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]

However, on the disk, moment of inertIa is:

[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]

The time t = 10s

∴

Frictional torques by the wall on the disk is:

[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]

Finally, the frictional force is calculated as:

[tex]F = \dfrac{T}r{}[/tex]

[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]