A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?

Answer:

The photosphere is the visible "surface" of the sun. So your answer would be C.

Explanation: its right

Answer:

239 rpm

Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:

100πcm/rev

75,000cm/min

​750 min rev≈238.7324146RPM

Answer:

A. Increases

Explanation:

As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.

Answer:

It decreases so it is B

Explanation:

As altitude rises, air pressure drops.

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m)  = .218 / m^2

x =  .467 m

Answer:

because they are found freely in nature uncombined so they are highly reactive with other elements

Answer:

 K_f = 1881.6 J

Explanation:

To solve this exercise, let's start by finding the velocities of the bodies.

We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved

initial instant. Before the explosion

        p₀ = M v₀

final instant. After the explosion

        p_f = m₁ v + m₂ 0

the moeoto is preserved

         p₀ = p_f

         M v₀ = m₁ v

         v = [tex]\frac{m_1}{M}[/tex]  v₀

in the exercise they indicate that the most massive part has twice the other part

         M = m₁ + m₂

         M = 2m₂ + m₂ = 3 m₂

         m₂ = M / 3

so the most massive part is worth

        m₁ = 2 M / 3

we substitute

        v = ⅔ v₀

with the speed of each element we can look for the kinetic energy

initial

         K₀ = ½ M v₀²

Final

         K_f = ½ m₁ v² + 0

         K_f = ½ (⅔ M) (⅔ v₀)²

         K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)

         K_f = [tex]\frac{8}{27}[/tex]  K₀

the energy added to the system is

         ΔK = Kf -K₀

         ΔK = (8/27 - 1) K₀

         ΔK = -0.7 K₀

         K_f = K₀ + ΔK

         K_f = K₀ (1 -0.7)

         K_f = 0.3 K₀

let's calculate

         K_f = 0.3 (½ 64 14²)

         K_f = 1881.6 J

According to Coulomb's law, the force between the given charges is 0.225N which is explained below.

Force on two identical charges q separate by a distance of r is given by:

F = kq²/r²

where k is Coulomb's constant

q is the charge

r is the separation between the charges

Given that q = 1×10⁻⁵C,

and r = 2m

So, the force between the given charges will be:

F = (9×10⁹)(1×10⁻⁵)²/2²

F = 0.225N is the required force.

Learn more about Coulomb's law:

https://brainly.com/question/506926?referrer=searchResults

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Answer:

1 W = 1 J / sec       Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min    energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

Answer:

Most likely (B)

Explanation:

B in the passage is the most representative out of all your choices and it has evidence from the passage  

Hope dis helps Jit!

Sorry i forgot to type C

B and C both measure mass while the others are calculations and are bias

The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.

When  a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.

When  a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.

Learn more about  gravitational field here:

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#SPJ2

Answer:

B. 3.1 × 10^5 V

Explanation:

Answer:

B

Explanation:

e2021

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

Answer:

5.48 m/s.

Explanation:

Use the formula a=v^2/r.

Answer:

0.54 sec

Explanation:

Answer:

Explanation:

Speed = 14.2 m/s

Distance = 514 m

Time = Distance / Speed

Time = 514 / 14.2

Time = 36.19 seconds

Answer:

#1 - True (touch) charging when electric conductors actually touch one another.

#2. True - electrical charges are conserved (not destroyed)

Answer:

[tex]1.62\times 10^{-8}\ \text{s}[/tex]

Explanation:

[tex]\epsilon_0[/tex] = Vacuum permittivity = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

[tex]A[/tex] = Area = [tex]10\times 2\times 10^{-4}\ \text{m}^2[/tex]

[tex]d[/tex] = Distance between plates = 1 mm

[tex]V_c[/tex] = Changed voltage = 60 V

[tex]V[/tex] = Initial voltage = 100 V

[tex]R[/tex] = Resistance = [tex]1000\ \Omega[/tex]

Capacitance is given by

[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}[/tex]

We have the relation

[tex]V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}[/tex]

The time taken for the potential difference to reach the required level is [tex]1.62\times 10^{-8}\ \text{s}[/tex].

Answer:

1. Wavelength = 3.2 m

2. Amplitude = 0.6 m

Explanation:

1. Determination of the wavelength.

The wavelength of a wave is defined as the distance between two successive crest. This implies that for every complete vibration, there is one wavelength.

From the diagram given above, we can see that the wave makes 2½ vibrations.

This means that there are 2½ equal wavelength of the wave. Therefore, the wavelength can be obtained as follow:

Length (L) = 8 m

Wavelength (λ) =?

2½ λ = L

5/2 λ = 8

5λ / 2 = 8

Cross multiply

5λ = 2 × 8

5λ = 16

Divide both side by 5

λ = 16 / 5

λ = 3.2 m

Therefore, wavelength of the wave is 3.2 m.

2. Determination of the amplitude.

The amplitude of a wave is defined as the maximum displacement of the wave from the origin.

From the diagram given above, the distance between the maximum and minimum displacement is given as 1.2 m. Thus, we can obtain the amplitude of wave as follow:

Distance between the maximum and minimum displacement (D) = 1.2

Amplitude (A) =?

A = ½D

A = ½ × 1.2

A = 0.6 m

Thus, the amplitude of the wave is 0.6 m

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Explanation:

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

Answer:

D:#2 African Plate

Explanation:

The African Plate is flanked by the Red sea rift and the minor African plate.

The Red sea rift is a small part of a greater line of rifts known as the Great African Rift Valley. The rift valley is making several small lakes all through Africa and it will eventually split up the African continent.

The Red sea lift is the divergent boundary between the African plate and the Arabian plate. It means that the two plates are moving apart or spreading apart.

Answer:

[tex]0.842\ \text{lb ft}[/tex]

[tex]0.1052\ \text{lb ft}[/tex]

Explanation:

d = Diameter of wheel = 6 in

r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]

t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]

w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]

[tex]t_2[/tex] = Time taken to slow down = 35 s

[tex]t_1[/tex] = Time taken to reach operating speed = 5 s

[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

Weight is given by

[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]

Mass is given by

[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]

Moment of inertia is given by

[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]

Angular acceleration while slowing down is given by

[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]

Frictional moment is given

[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]

Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]

Angular acceleration while speeding up is given by

[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]

Motor torque is given by

[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]

Motor torque is [tex]0.842\ \text{lb ft}[/tex].

Answer:

a = 52s²

Explanation:

How to find acceleration

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Solve

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

Answer:

Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.

Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.

Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].

In other words:

[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].

[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Combine the two equations (equate the right-hand side) to obtain:

[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:

[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].

[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].