Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If the period of the first planet P1 is 750 years what is the mass, in kg, of the star it orbits around?

Answer:

Explanation:

This is an excellent question to get an answer for. It teaches you much about the nature of physics.

The answer is no.

The distance will be quite different. The time might be different in getting to the distance.  But the acceleration will be the same in either case.

How do you know? Look at one of the formulas, say

d = vi * t + 1/2*a * t^2

What does vi do? vi will alter both t and d. if vi = 0 then both d and/or t will be found. But what will "a" do? Is there anything else acting in the up or down line of action? You should answer no.

If vi is not zero, t will be less and d will take less time to get where it is going.

Yes, the initial velocity of an object affects its acceleration. If you drop an object, the initial velocity is zero, resulting in a constant acceleration solely due to gravity. If you throw the object downward, the initial velocity adds to the acceleration due to gravity, leading to a higher overall acceleration.

Yes, the initial velocity of an object can affect its acceleration. When you drop an object, its initial velocity is zero, so the acceleration is solely due to gravity and is constant at approximately 9.8 m/s2 (assuming no air resistance). However, if you throw an object downward, it already has an initial downward velocity, which adds to the acceleration due to gravity. So, the acceleration of the object after you release it will be greater than if you had dropped it.

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Answer:

a) 1.34 Volts

b) 3.08 W

c) 2.68 W

Explanation:

Given:

Emf of the cell, E = 1.54 V

current, i = 2.0 A

internal resistance, r = 0.100Ω

(a) Terminal voltage (V) = E - v

where,

v is the potential difference across the resistance 'r'

now,

according to the Ohm's Law, we have

v = i × r  

substituting the values in the above equation we get

v = 2.0 × 0.100 = 0.2 Volts

thus,

Terminal voltage (V) = (1.54 - 0.2) = 1.34 V

(b) Now, the Total power (P) is given as

P = E × i = (1.54 × 2.0) = 3.08 W  

(c) Power into its load = [terminal voltage, v] * i  

= (1.34 × 2.0) = 2.68 W

Final answer:

Terminal voltage of a carbon-zinc cell is 1.34 V, producing 2.68 W of electrical power, with 2.68 W going to the load.

Explanation:

Terminal voltage: The formula to calculate terminal voltage is V = EMF - I * r, where V is the terminal voltage, EMF is the electromotive force, I is the current, and r is the internal resistance of the cell. So, V = 1.54 V - 2.00 A * 0.100 Ω = 1.34 V.

Electrical power: The electrical power produced by the cell is given by P = V * I, where P is power, V is voltage, and I is current. Substituting the values, P = 1.34 V * 2.00 A = 2.68 W.

Power to load: The power delivered to the load is equal to the voltage supplied to the load times the current flowing through it. Hence, the power to the load is P = V_load * I = 1.34 V * 2.00 A = 2.68 W.

Answer:

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Explanation:

Time period of simple pendulum is given by

         [tex]T=2\pi\sqrt{\frac{l}{g}}[/tex], l is the length of pendulum, g is acceleration due to gravity value.

We can solve acceleration due to gravity as

            [tex]g=\frac{4\pi^2l}{T^2}[/tex]

Here

  Length of pendulum = 1.20 m

  Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.

  Period, [tex]T=\frac{450}{100}=4.5s[/tex]

Substituting

         [tex]g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2[/tex]

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

[tex]m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}[/tex]

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2  = 0 m/s

putting all value in the above equation

[tex]1.14 *2.0+ 0 =(1.14+1.14)v^{2}[/tex]

[tex]v =\frac{1.14*2.0}{1.14+1.14}[/tex]

v = 1 m/s

By applying the conservation of momentum principle, the speed of the two blocks immediately after the collision is found by equating the momentum before and after the collision.

In this physics problem, we need to determine the speed of the two blocks attached by a spring immediately after the collision. The principle of conservation of momentum can be applied to this. Before the collision, only block m1 is moving, so the total momentum is m1*v1. After the collision, the two blocks move together at speed v', resulting in total momentum of (m1 + m2) * v'. According to the principle of conservation of momentum, the momentum before collision equals to momentum after collision. Hence, m1*v1 = (m1 + m2)*v'.

Solving this equation will yield the value of v'. Substituting the values of m1 = 2.27 kg, m2 = 1.14 kg, and v1 = 2.00 m/s will result in v' = (2.27 kg * 2.00 m/s) / (2.27 kg + 1.14 kg), which gives the speed of the blocks immediately after collision.

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Answer:

The average force exerted on the bullet are of F=9000 Newtons.

Explanation:

t= 2*10⁻³ s

m= 0.03 kg

V= 600 m/s

F*t= m*V

F= (m*V)/t

F= 9000 N

The average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N). when a bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gunpowder.

Given:

Mass of the bullet (m) = 0.0300 kg

The final velocity of the bullet (v) = 600 m/s

Time taken to reach the final velocity (t) = 2.00 ms = 2.00 × 10⁻³ s

acceleration (a) = (change in velocity) / (time taken)

a = (v - u) / t

a = (600 - 0 ) / (2.00 × 10⁻³)

Now, we can calculate the average force using Newton's second law:

force (F) = mass (m) × acceleration (a)

F = 0.0300 × [(600 ) / (2.00 × 10⁻³)]

F = 0.0300× (3.00 × 10⁵)

F = 9000 N

Therefore, the average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N).

To know more about the force exerted:

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Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

[tex]PV=nRT[/tex]

For a gas

[tex]P_{1}V_{1}=nRT_{1}[/tex]

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

[tex]10\times V=nR\times286[/tex]....(I)

When the temperature of the gas is increased

Then,

[tex]P_{2}V_{2}=\dfrac{n}{2}RT_{2}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}[/tex]

[tex]\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}[/tex]

[tex]P_{2}=\dfrac{10\times368}{2\times286}[/tex]

[tex]P_{2}= 6.433\ atm[/tex]

[tex]P_{2}=6.4\ atm[/tex]

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

Answer:

The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

Explanation:

Given that,

Wavelength of Raman line [tex]\lambda'=4768.5\ A[/tex]

Wavelength [tex]\lambda=4358.3\ A[/tex]

We need to calculate the frequency

Using formula of frequency

[tex]f =\dfrac{c}{\lambda}[/tex]

For 4748.5 A

The frequency is

[tex]f'=\dfrac{3\times10^{8}}{4748.5\times10^{-10}}[/tex]

[tex]f' =6.32\times10^{14}\ Hz[/tex]

For 4358.3 A

The frequency is

[tex]f=\dfrac{3\times10^{8}}{4358.3\times10^{-10}}[/tex]

[tex]f=6.88\times10^{14}\ Hz[/tex]

We need to calculate the shift

[tex]\Delta f=f-f'[/tex]

[tex]\Delta f=(6.88-6.32)\times10^{14}\ Hz[/tex]

[tex]\Delta f=0.56\times10^{14}\ Hz[/tex]

Hence, The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

Answer:

Object 1 is hotter.

Explanation:

Object 1 T=[tex]0^{\circ}C[/tex]  

Object 2 T=0 K

Object 3 T=[tex]0^{\circ}F[/tex]

Relation between Celcius ,Kelvin and Fahrenheit

[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}[/tex]

K=C+273,  [tex]K=\dfrac{5}{9}(F-32)+273[/tex].

So now we will convert all in one unit.

Object 1 T=273 K

Object 2 T=0 K

Object 3 T=255.22 K

From above we can say that Object 2 is coolest and object 1 is hottest.

So  Object 1 is hotter.

Explanation:

Charge of electron in He, [tex]q_e=1.6\times 10^{-19}\ kg[/tex]

Charge of proton in He, [tex]q_p=1.6\times 10^{-19}\ kg[/tex]

Distance between them, [tex]d=2.7\times 10^{-10}\ m[/tex]

We need to find the electric force between them. It is given by :

[tex]F=k\dfrac{q_eq_p}{d^2}[/tex]

[tex]F=-9\times 10^9\times \dfrac{(1.6\times 10^{-19}\ C)^2}{(2.7\times 10^{-10}\ m)^2}[/tex]

[tex]F=-3.16\times 10^{-9}\ N[/tex]

Since, there are two protons so, the force become double i.e.

[tex]F=2\times 3.16\times 10^{-9}\ N[/tex]

[tex]F=6.32\times 10^{-9}\ N[/tex]

So, the correct option is (c). Hence, this is the required solution.

Answer:

Explanation:

A proton is positively charged in nature.

Let an electric field strength is E.

The force on a charged particle placed in an electric field is given by

F = q x E

Where q is the charge

Here, the proton is positively charged, so the direction of force acting on the proton is same as teh direction of electric field strength.

Thus, the motion of proton is parallel to the electric field and the motion is accelerated.

Answer:

Change in ball's momentum is 1.5 kg-m/s.

Explanation:

It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.15\ kg(-3.5\ m/s-6.5\ m/s)[/tex]

[tex]\Delta p=-1.5\ kg-m/s[/tex]

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

The change in the magnitude of the ball's momentum is 1.125 kg·m/s.

The change in the magnitude of an object's momentum can be calculated by subtracting its initial momentum from its final momentum. In this case, the initial momentum of the ball is calculated by multiplying its mass (0.15 kg) by its initial speed (6.5 m/s), and the final momentum is calculated by multiplying its mass by its final speed (3.5 m/s).

So, the change in the magnitude of the ball's momentum is calculated as:

|pf| - |pi| = (0.15 kg)(3.5 m/s) - (0.15 kg)(6.5 m/s) = -0.15 kg·m/s - 0.975 kg·m/s = -1.125 kg·m/s.

Since momentum is a vector quantity, the change in its magnitude is given by its absolute value, which is 1.125 kg·m/s. Therefore, the correct answer is A. 1.125 kg·m/s.

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Answer:

The number of atoms are [tex]1.86\times10^{8}[/tex].

Explanation:

Given that,

Diameter [tex]D = 1.40\times10^{2}\ pm[/tex]

[tex]D=1.40\times10^{2}\times10^{-12}\ m[/tex]

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

[tex]Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}[/tex]

[tex]Number\of\atoms =1.86\times10^{8}[/tex]

Hence, The number of atoms are [tex]1.86\times10^{8}[/tex].

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance [tex]r= 5.00\ \Omega[/tex]

(a). We need to calculate the current

Using rule of loop

[tex]E-IR-Ir=0[/tex]

[tex]I=\dfrac{E}{R+r}[/tex]

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

[tex]I=\dfrac{3.200}{1.00\times10^{3}+5.00}[/tex]

[tex]I=3.184\times10^{-3}\ A[/tex]

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

[tex]V=E-Ir[/tex]

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

[tex]V=3.200-3.184\times10^{-3}\times5.00[/tex]

[tex]V=3.18\ V[/tex]

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

[tex]\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }[/tex]

[tex]\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}[/tex]

Hence, This is the required solution.

The current flowing in the circuit is 3.195 milliamps. The terminal voltage is calculated to be 2.984V. The ratio of the terminal voltage to the emf is 0.9325 which shows that the terminal voltage is 93.25% of the emf due to the voltage drop resulted from the internal resistance of the battery.

The question is about the terminal voltage and internal resistance of a battery. To answer this, first we need to understand that terminal voltage is the potential difference (voltage) between two terminals of a battery, and it's slightly less than the emf due to internal resistance of the battery.

(a) The current (I) in the circuit is found using Ohm’s law: I = emf / (R_load + r) = 3.200V / (1.00kΩ + 5.00Ω) = 3.195 milliamps.

(b) The terminal voltage (V) is calculated by: V = emf - Ir = 3.200V - (3.195mA * 5.00Ω) = 2.984V.

(c) The ratio of the measured terminal voltage to the emf is V / emf = 2.984V / 3.200V = 0.9325. This shows that the terminal voltage is 93.25% of the emf, which accounts for the voltage drop due to the internal resistance of the battery.

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Answer:

Answer to the question:

Explanation:

A black hole is a finite region of space within which there is a mass concentration high and dense enough to generate a gravitational field such that no material particle, not even light, can escape it.

Answer:

[tex]Q = 1.45 \times 10^{-7} C[/tex]

Explanation:

Here flux passing through each surface is given

so total flux through whole cube is given by

[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]

[tex]\phi = 16400 Nm^2 C[/tex]

now we also know that total flux through a closed surface depends on the total charge enclosed in the surface

So we will have

[tex]\frac{Q}{\epsilon_0} = 16400[/tex]

[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]

[tex]Q = 1.45 \times 10^{-7} C[/tex]

The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:

[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]

where:

- [tex]\( \Phi \)[/tex] is the electric flux,

- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,

- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].

For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].

[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]

Now, we can set up an equation using the given values:

[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]

[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]

Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:

[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]

[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]

[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]

So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

Answer:

Part a)

a = 1.62 m/s/s

Part b)

a = 3.70 m/s/s

Explanation:

Part A)

Acceleration due to gravity on the surface of moon is given as

[tex]a = \frac{GM}{R^2}[/tex]

here we know that

[tex]M = 7.35 \times 10^{22} kg[/tex]

[tex]R = 1.74 \times 10^6 m[/tex]

now we have

[tex]a_g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}[/tex]

[tex]a_g = 1.62 m/s^2[/tex]

Part B)

Acceleration due to gravity on surface of Mercury is given as

[tex]a = \frac{GM}{R^2}[/tex]

here we know that

[tex]M = 3.30 \times 10^{23} kg[/tex]

[tex]R = 2.44 \times 10^6 m[/tex]

now we have

[tex]a_g = \frac{(6.67 \times 10^{-11})(3.30 \times 10^{23})}{(2.44 \times 10^6)^2}[/tex]

[tex]a_g = 3.70 m/s^2[/tex]

The distance traveled by a particle with velocity function

over the first t seconds can be obtained by integrating that function from 0 to t using integration by parts.

The distance traveled by a particle can essentially be obtained by integrating the velocity function over a given timeframe. In this case, the velocity function is given by

To calculate the distance traveled over the first t seconds, we need to integrate this function from 0 to t. Therefore, the integral ∫v(t)dt from 0 to t will provide the required solution. Use the method of integration by parts, choosing u=t^2 and

then complete the integration process following the standard procedure.

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im not that smart but maybe 2 x the mass of the wieght will give u the answer

Explanation:

It is given that,

Mass of the metal wire, m = 500 g = 0.5 kg

Tension in the wire, T = 80 N

Length of wire, l = 50 cm = 0.5 m

(a) The speed of the transverse wave is given by :

[tex]v=\sqrt{\dfrac{T}{M}}[/tex]

M is the mass per unit length or M = m/l

[tex]v=\sqrt{\dfrac{T.l}{m}}[/tex]

[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{0.5\ kg}}[/tex]

v = 8.94 m/s

(b) If the wire is cut in half, so l = l/2

[tex]v=\sqrt{\dfrac{T.l}{2m}}[/tex]

[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{2\times 0.5\ kg}}[/tex]

v = 6.32 m/s

Hence, this is the required solution.

Answer:

299034 N/m²

Explanation:

1 lbs = 4.448 N

1 in = 0.0254 m

1 in² = 0.254² m²

thus,

[tex]1\frac{lbs}{in^2} = \frac{4.448N}{0.0254^2m^2}=6894.413N/m^2[/tex]

therefore,

43.36lbs/in² in N/m² will be

= 43.36 × 6894.413

= 298941.77  N/m² ≈ 299034 N/m²

so the correct option is 299034 N/m²

Final answer:

The increase in temperature of the water is 0.796 °C.

Explanation:

To calculate the increase in temperature of the water, we first need to calculate the total energy absorbed by the collector. The power incident on the collector can be calculated by multiplying the solar radiation intensity by the collector area:

Power incident on the collector = 1000 W/m² × 4 m² = 4000 W

The energy absorbed by the collector can be calculated by multiplying the power incident on the collector by the conversion efficiency:

Energy absorbed by the collector = 4000 W × 0.25 = 1000 J/s

Now we can calculate the increase in temperature of the water using the specific heat formula:

ΔT = Energy absorbed by the collector / (mass of water × specific heat of water)

ΔT = 1000 J/s / (30 kg × 4186 J/kg×°C) = 0.796 °C

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          [tex]t=\frac{200}{9.81}=20.38s[/tex]

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

The radii of curvature can be determined using the Lens Maker's Equation. For a planar-convex lens, we can consider one surface as flat and another as curved. If the refractive index decreases, the radius of curvature would increase.

To find the radii of curvature for the planar-convex thin lens in air, you can use the Lens Maker's Equation, which is 1/f = (n-1)(1/R1 - 1/R2). Here, f is the focal length, n is the refractive index of the glass, R1 and R2 are the radii of curvature for the two surfaces of the lens.

For a planar-convex lens, one surface is flat (which is the planar side) and another surface is curved (which is the convex side). So, we can consider R1 = ∞ for the flat surface and R2 = R (the required radius) for the convex surface. By substituting these values into the Lens Maker's Equation, we can solve for the radius of curvature of the convex surface.

If n was reduced to 1.500, the radius of curvature would increase because, according to the Lens Maker's Equation, radius of curvature is inversely proportional to (n-1). Thus, as n decreases, the radius of curvature increases.

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Answer:

A) 50.0 g Al

Explanation:

We can calculate the temperature change of each substance by using the equation:

[tex]\Delta T=\frac{Q}{mC_s}[/tex]

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

[tex]C_s[/tex] is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

[tex]\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C[/tex]

B) m = 50.0 g, Cs = 0.385 J/g°C

[tex]\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C[/tex]

C) m = 25.0 g, Cs = 0.79 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C[/tex]

D) m = 25.0 g, Cs = 0.128 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C[/tex]

E) m = 25.0 g, Cs = 0.235 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C[/tex]

As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.

The substance that would show the smallest temperature change upon gaining 200.0 J of heat is Au (Gold), as calculated using the formula for calculating heat (Q = mcΔT) and rearranging for ΔT, then substituting the given values.

The substance that would show the smallest temperature change upon gaining 200.0 J of heat can be determined using the formula used to calculate heat (Q), which is Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change. We want to find the smallest temperature change, so we rearrange the equation to solve for ΔT, which gives us ΔT = Q/(mc). By substituting the given values for each substance into this equation, we find that the smallest temperature change is for Au (Gold).

For Au: ΔT = 200.0J / (25.0g x 0.128 J/g°C) = 62.5°C. All other substances have a smaller temperature change when they absorb 200.0J of heat, due to their higher specific heat capacity.

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Answer:

9.66 x 10^-6 m

Explanation:

Use the Wein's displacement law

[tex]\lambda _{m}\times T = b[/tex]

Where, b is the Wein's constant

b =  2.898 x 10^-3 meter-kelvin

So, λm x 300 =  2.898 x 10^-3

λm = 9.66 x 10^-6 m

Answer:

The normal force acting on the block is 980 N.

Explanation:

Mass of box = 100 kg

Coefficient of static friction = 0.40

Coefficient of kinetic friction = 0.20

We need to calculate the normal force

We know that,

When the object is placed on the flat surface then the normal force is equal to the weight of the object.

So. The normal force = mass x acceleration due to gravity

[tex]N=mg[/tex]

Where, N = Normal force

m = mass of object

g = acceleration due to gravity

[tex]N=100\times9.8[/tex]

[tex]N=980\ N[/tex]

Hence, The normal force acting on the block is 980 N.

Final answer:

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N.

Explanation:

The question asks about the normal force acting on a 100-kg box at rest on the floor. To find the normal force on the box, we can use the equation that relates the normal force to the weight of the object, which is W = mg, where m is the mass and g is the acceleration due to gravity.

Given that the mass of the box is 100 kg and the acceleration due to gravity is 9.80 m/s², the normal force can be calculated as:

Normal Force (N) = Mass (m) × Acceleration due to gravity (g)

Normal Force = 100 kg × 9.80 m/s² = 980 N

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N. It is often confused with the applied force P, but the normal force is related to the box's weight, not the horizontal force applied.

Answer:

3.08 Nm

Explanation:

N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree

The angle made with the normal of the coil, theta = 90 - 30 = 60 degree

Torque = N I A B Sin Theta

Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60

Torque = 3.08 Nm

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        [tex]\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}[/tex]

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

[tex]a=\frac{11.7-26.5}{5.2}=-2.85m/s^2[/tex]

Angular acceleration

        [tex]\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2[/tex]

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

Answer:

0.081 m /s

Explanation:

According to the conservation of momentum, the momentum of a system is conserved when no external force is applied on a body.

momentum of the system before throwing = momentum of the system after throwing

Let v be the velocity of quarterback after throwing the football.

80 x 0 + 0.43 x 0 = 80 x v + 0.43 x 15

0 = 80 v + 6.45

v = - 0.081 m /s

The negative sign shows that he is moving in backward direction.

The quarterback will be moving backward at approximately 0.080625 m/s just after releasing the football due to the conservation of momentum.

The student's question involves practicing Problem-Solving Strategy 11.1 for conservation of momentum problems. Specifically, the student is asked to calculate how fast an 80-kg quarterback will be moving backward just after releasing a football that has a mass of 0.43 kg and is thrown horizontally at a speed of 15 m/s. To solve this, we use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.

Before the throw, the total momentum of the system (quarterback and football) is zero because both are stationary with respect to horizontal motion. After the throw, the combined momentum must still be zero. We can set up the equation mQB × vQB + mball × vball = 0, where mQB is the mass of the quarterback, vQB is the quarterback's velocity after the throw, mball is the mass of the football, and vball is the velocity of the football. This simplifies to vQB = -(mball × vball) / mQB.

Plugging in the numbers gives us vQB = -(0.43 kg × 15 m/s) / 80 kg, which yields a velocity of vQB ≈ -0.080625 m/s. The negative sign indicates that the quarterback is moving in the opposite direction of the throw, which is backward.

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The magnitude of the angular momentum of the paperclip attached to the spinning vinyl record is approximately 0.0033 kg m²/s. This is calculated using the formulas for moment of inertia and angular momentum.

To calculate the angular momentum of the paperclip, we first need to know the moment of inertia (I) of the paperclip. The moment of inertia can be calculated using the formula I = mR² where 'm' is the mass of the paperclip (converted into kg - 0.018 kg) and 'R' is the radius of the record player (converted into m - 0.24 m).

So, I = 0.018 kg * (0.24 m)² = 0.0010368 kg m².

Next, we use the formula for angular momentum (L), which is L = Iω, where ω is the angular velocity. Given ω = 3.2 rad/s, we plug these values into our formula:

L = 0.0010368 kg m² * 3.2 rad/s = t0.00331776 kg m²/s.

Thus, rounding to the nearest ten-thousandth, the magnitude of the angular momentum of the paperclip is 0.0033 kg m²/s.

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The magnitude of the angular momentum of the paper clip is approximately [tex]\( 0.0033 \text{ kg m}^2/\text{s} \)[/tex].

The magnitude of the angular momentum of the paper clip is given by the formula [tex]\( L = I\omega \)[/tex], where I is the moment of inertia of the paper clip and [tex]\( \omega \)[/tex] is the angular velocity of the record.

Given:

- Mass of the paper clip, [tex]\( m = 18 \) g \( = 0.018 \)[/tex] kg (after converting grams to kilograms)

- Diameter of the record, [tex]\( d = 48 \) cm \( = 0.48 \)[/tex] m (after converting centimeters to meters)

- Radius of the record, [tex]\( r = \frac{d}{2} = \frac{0.48}{2} = 0.24 \)[/tex]m

- Angular velocity, [tex]\( \omega = 3.2 \)[/tex] rad/s

Now, we calculate the moment of inertia I:

[tex]\[ I = mr^2 = 0.018 \times (0.24)^2 \] \[ I = 0.018 \times 0.0576 \] \[ I = 0.0010368 \text{ kg m}^2 \][/tex]

Next, we calculate the angular momentum L:

[tex]\[ L = I\omega \] \[ L = 0.0010368 \times 3.2 \] \[ L = 0.00331776 \text{ kg m}^2/\text{s} \][/tex]

Rounding to the nearest ten-thousandth, we get:

[tex]\[ L \approx 0.0033 \text{ kg m}^2/\text{s} \][/tex]