In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

Answer:

: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

Yes. Because it opposes the law of friction

I hope this helps.

Explanation:

Please mark me brainliest

Answer: hello tables and data related to your question is missing attached below are the missing data

answer:

a) I = I₁ = I₂ = I₃ = 0.484 mA

b) I₁ =  0.016 amps

   I₂ =  0.0016 amps

   I₃ = 7.27 * 10^-4 amps

c)  I₁ = 1.43 * 10^-3 amp

    I₂ =  0.65 * 10^-3 amps

Explanation:

A) magnitude of current for Part 1

Resistors are connected in series

Req = r1 + r2 + r3

       = 3300 Ω  ( value gotten from table 1 ) ,

          V = 1.6 V ( value gotten from table )

hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA

The magnitude of current is the same in the circuit

Vi = I * Ri

B) magnitude of current for part 2

Resistors are connected in parallel

V = 1.6 volts

Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) +  R3 ]

      = [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]

      = 87.30 Ω

For a parallel circuit the current flow through each resistor is different

hence the magnitude of the currents are

I₁ = V / R1 = 1.6 / 100 = 0.016 amps

I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps

I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps

C) magnitude of current for part 3

Resistors are connected in combination

V = 1.6 volts

Req = R1 + ( R2 * R3 / R2 + R3 )

       = 766.66 Ω

Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps

magnitude of currents

I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps

I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps

Answer:

1.67 m

Explanation:

The potential energy change of the small ball ΔU equals the work done by the buoyant force, W

ΔU = -W

Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5

ΔU = mgΔh

ΔU = ρVgΔh

Now, W = ρ'VgΔh'   where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h

So, ΔU = -W

ρVgΔh = -ρ'VgΔh'

ρVg(h - 5) = -ρ'Vgh

ρ(h - 5) = -ρ'h

Since the density of the small ball equals 1/2 the density of water,

ρ = ρ'/2

ρ(h - 5) = -ρ'h

(ρ'/2)(h - 5) = -ρ'h

ρ'(h - 5)/2 = -ρ'h

(h - 5)/2 = -h

h - 5 = -2h

h + 2h = 5

3h = 5

h = 5/3

h = 1.67 m

So, the maximum depth the ball reaches is 1.67 m.

Complete question is;

A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:

+5q; 0

-6q; +2q

+2q; -3q

-4q; +12q

Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:

a. 2, 3, 1, 4

b. 1, 2, 3, 4

c. 2, 4, 3, 1

d. 1, 3, 4, 2

Answer:

c. 2, 4, 3, 1

Explanation:

In this question, we can say that;

q_in = q_b

Where;

q_in is the charge on the inner surface of the shell

q_b is the point charge on the shell.

Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.

Thus;

- For +5q; 0:

q_in = -(+5q)

q_in = -5q

- For -6q; +2q :

q_in = - (-6q)

q_in = +6q

- For +2q; -3q :

q_in = -(+2q)

q_in = -2q

- For -4q; +12q:

q_in = -(-4q)

q_in = +4q

Ranking the most positive to the least positive ones, we have;

+6q, +4q, -2q, -5q

This corresponds to options;

2, 4, 3, 1

Answer:

p = 1.2 kg-m/s

Explanation:

The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".

Mass of the object, m = 0.5 kg

Acceleration of the object, a = 0.3 m/s²

We need to find the momentum after 8 seconds.

We know that,

[tex]p=F\times t[/tex]

i.e.

p = mat

So,

[tex]p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s[/tex]

So, the momentum of the object is 1.2 kg-m/s.

Answer:

B . energy cannot be created or destroyed

Answer:

multiply mp and c^2

Explanation:

e=mc^2

Answer:

76rsfy7zfyuutfzufyztudzutdT7dFy9y8fr6s

Explanation:

rshyyjfshfsgfshfsyhrsyhuydtufhr6ra6yris7toe7r9w7rr6w996ryrowosotusuogsuoufsutot

Answer:

proton have higher energy than electron

Explanation:

tag me brainliest

Answer:

proton

Explanation:

proton is higher energy than the electron

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

Answer:

[tex]W=2KJ[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=50kg[/tex]

Initial Velocity [tex]v_1=8m/s[/tex]

Final Velocity [tex]v_2=12m/s[/tex]

Generally the equation for Work-done is mathematically given by

W=\triangle K.E

Therefore

 [tex]W=0.5M(v_2^2-v_1^2)[/tex]

 [tex]W=0.5*50(12^2-8^2)[/tex]

 [tex]W=2KJ[/tex]

Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

Let the radius is r. And outer is R.

Use the formula of current density

[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

[tex]F = BIl \ sin(\theta)[/tex]

(a) When the angle, θ = 60 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N[/tex]

(b) When the angle, θ = 90 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N[/tex]

(c) When the angle, θ = 120 ⁰

[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N[/tex]

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Answer:

Myopia curvature of the cornea if it is negative the curvatures are positive,

hypermetry,

Explanation:

Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative

therefore the curvature must decrease

To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase

Answer:

fair palybtgshsisuehdh

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]

Explanation:

The initial kinetic energy [tex]KE_0[/tex] is

[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]

When its mass and velocity are doubled, its new kinetic energy KE is

[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]

[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]

Therefore the kinetic energy will increase by a factor of 8.

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

Answer:

Approximately [tex]0.97\; \rm N[/tex]. This force would point away from the center of the square (to the left at [tex]45^\circ[/tex] above the horizontal direction.)

Explanation:

Coulomb's constant: [tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].

By Coulomb's Law, the electrostatic force between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] that are separated by [tex]r[/tex] (vacuum) would be:

[tex]\displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}[/tex].

Consider the charge on the top-left corner of this square.

Apply Coulomb's Law to find the electrostatic force between this charge and the charge on the lower-left corner.

Convert quantities to standard units:

[tex]q_1 = q_2 = 3 \times 10^{-6}\; \rm C[/tex].

[tex]r = 0.40\; \rm m[/tex].

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}[/tex].

As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.

Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately [tex]0.506\; \rm N[/tex]) that points leftwards.

For the charge to the lower-right of the top-left charge, [tex]r = \sqrt{2} \times 0.40\; \rm m[/tex]. Therefore:

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}[/tex].

This force would point to the top-left of the top-left charge, which is [tex]45^\circ[/tex] above the horizontal direction. Decompose this force into two components that are normal to one another:

Consider the net force on the top-left charge in two components:

Combine these two components to find the magnitude of the net force on this charge:

[tex]\begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}[/tex].

This force would point to the top-left of this charge (also at [tex]45^\circ[/tex] above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=[tex]\rho[/tex]

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

[tex]Density=\frac{mass}{volume}[/tex]

[tex]Mass=Density\times volume=Constant[/tex]

Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]

Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]

Using the formula

[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]

[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]

[tex]\rho_2=8\rho[/tex]

Hence, the density of  the compressed sphere=[tex]8\rho[/tex]

Option a is correct.

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• m = 1 kg:

∑ F (parallel) = mg sin(45°) - T = ma … … … [1]

∑ F (perpendicular) = n - mg cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• m = 0.4 kg:

∑ F (vertical) = T - mg = ma … … … [2]

Adding equations [1] and [2] eliminates T, so that

((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a

(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a

==>   a ≈ 2.15 m/s²

The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

Answer:

The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)

Second skater

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)

Where:

[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.

[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.

[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.

[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.

[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.

[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.

If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:

By (1):

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]

[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]

[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]

[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]

[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]

By (2):

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]

[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]

[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]

[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]

Answer:

Toaster = I = 14.67 A

Frying Pan = 11.83 A

Lamp = 0.71 A

Explanation:

The electric power is given as:

[tex]P = VI\\\\I = \frac{P}{V}[/tex]

where,

I = current

P = Power

V = Voltage = 120 V

FOR TOASTER:

P = 1760 W

Therefore,

[tex]I = \frac{1760\ W}{120\ V}[/tex]

I = 14.67 A

FOR FRYING PAN:

P = 1420 W

Therefore,

[tex]I = \frac{1420\ W}{120\ V}[/tex]

I = 11.83 A

FOR LAMP:

P = 85 W

Therefore,

[tex]I = \frac{85\ W}{120\ V}[/tex]

I = 0.71 A