A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.

Answer: 4 A

Explanation:

Given

Voltage [tex]V=230\ V[/tex]

Power [tex]P=1\ kW[/tex]

Power is given by [tex]P=VI\\[/tex]

Insert the values

[tex]\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A[/tex]

The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.

Answer:

length =  7*10^(-8)km

width = 4.666*10^(-8) km

Explanation:

We know that:

1 μm = 1*10^(-6) m

and

1km = 1*10^3 m

or

1m = 1*10^(-3) km

if we replace the meter in the first equation, we get:

1 μm = 1*10^(-6)*1*10^(-3) km

1 μm = 1*10^(-6 - 3)km

1 μm = 1*10^(-9)km

Now with this relationship we can transform our measures:

Length: 70 μm is 70 times 1*10^(-9)km, or:

L = 70*1*10^(-9)km = 7*10^(-8)km

And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:

W = 46.66*1*10^(-9)km = 4.666*10^(-8) km

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi  =  r ω (radius X angular velocity)

 angular velocity, ωi = v/r

ωi = 26.389  m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

 = 37.879rad/s

 How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

Answer:

(a) the mass of the second body is 1.2 kg

(b)  the speed of the two-body center of mass 2.5 m/s

Explanation:

Given;

mass of the body, m₁ = 2 kg

let the mass of the second body = m₂

let the initial speed of the first body, = u₁ = 4 m/s

then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁

initial speed of the second body, u₂ = 0

let the final speed of the second body = v₂

Apply principle of conservation of linear momentum to determine the mass of the second body;

m₁u₁  +  m₂u₂ = m₁v₁  +  m₂v₂

2u₁  +  0(m₂)  =  2(0.25u₁)  +  m₂v₂

2u₁   =  0.5u₁  +  m₂v₂

1.5(4) = m₂v₂

6 = m₂v₂

Apply one-directional velocity

u₁ + v₁ = u₂ + v₂

u₁ + (0.25u₁) = 0 + v₂

1.25u₁ = v₂

1.25(4) = v₂

5 = v₂

Then, the mass of the second body is calculated as;

m₂v₂ = 6

5m₂ = 6

m₂ = 6/5

m₂ = 1.2 kg

(b)  the speed of the two-body center of mass after collision;

[tex]V_c_m = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\\\V_c_m = \frac{2(0.25\times 4) \ + \ 1.2(5)}{2\ + \ 1.2} \\\\V_c_m = 2. 5 \ m/s[/tex]

Answer:

0.4

Explanation:

[tex] \frac{1}{2} mv ^{2} [/tex]

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

Answer :

Considering initial velocity is 0,

It takes about 0.4 seconds.

Use the equation v-u =at

4-0 = 9.8×t

t = 4/9.8 = 0.4 seconds approximately.

Answer:

the length of the pipe is 0.85 m or 85 cm

Explanation:

Given the data in the question;

The successive harmonics are;  700 Hz , 900 Hz , and 1100 H

Now, for a closed pipe,

length of pipe (L) = λ/4

Harmonics; 1x, 3x, 5x, 7x, 9x, 11x

1100Hz - 900Hz = 200Hz

⇒ 2x = 200Hz

x = 100Hz ( fundamental frequency )

λ = V/f = 340 /100 = 3.4 m

Now

Length L = λ / 4

L = 3.4 / 4

L = 0.85 m or 85 cm

Therefore, the length of the pipe is 0.85 m or 85 cm

Answer:

The elastic potential energy of the spring is 0.28 J

Explanation:

Given;

mass of the block, m = 0.05 kg

spring constant, k = 350 N/m

extension of the spring, x = 4 cm = 0.04 m

The elastic potential energy of the spring is calculated as;

[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]

Therefore, the elastic potential energy of the spring is 0.28 J

Answer:

15.C Mucus reduces friction

Answer:

[tex]^{214} _{83} Bi[/tex]  → [tex]^{210}_{81}Tl[/tex] + [tex]^4_2He[/tex]

Explanation:

[tex]^{214} _{83} Bi[/tex]  → [tex]x[/tex] + [tex]^4_2He[/tex]

Subtract the [tex]^4_2He[/tex] from the [tex]^{214} _{83} Bi[/tex]

214 - 4 = 210

83 -2 = 81

Therefore, [tex]x[/tex] = [tex]^{210}_{81}Tl[/tex]

Answer:

W = 16.4 kJ

Explanation:

Given that,

There are 135 steps from the ground floor to the sixth floor.

Each step is 16.6 cm tall.

The mass of a person, m = 73.5 kg

We need to find the work done by the person. We know that,

Work done = Fd

Where

d is the displacement, d = 135 × 0.166 = 22.41

So,

W = 73.5 × 10 × 22.41

= 16471.35  J

or

W = 16.4 kJ

So, 16.4 kJ is the work done by the person.

Answer:

The answer is "[tex]16.79\ \frac{m}{s}[/tex]"

Explanation:

In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]

Using formula:

[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]

[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]

Answer:

Shiny surface.

Explanation:

We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.

Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.

This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,

Answer:

tq1=EoA/D

-2q1=-2EoAV/d

Option C.

Answer:

tq1=EoA/D

-2q1=-2EoAV/D

Explanation:

option C is correct

Answer:

D) The bicycle is not in motion.

Explanation:

Study the position-time graph for a bicycle.

Which statement is supported by the graph?

A) The bicycle has speed but not velocity.

B) The bicycle is moving at a constant velocity.

C) The bicycle has a displacement of 3 m.

D) The bicycle is not in motion.

Solution:

Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.

Speed is the time rate of change of distance. It is the ratio of distance to time taken.

From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.

From the position-time data provided, it can concluded that the bicycle is not in motion.

Motion of a body involves a change in the position of that body with time.

A body in motion is constantly changing position or orientation as time passes.

The body may move with constant velocity/speed or changes in its velocity.

A position-time graph provides information about the motion of a body.

From the data provided:

The position of the bicycle remains the same for all time intervals.

Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.

Learn more about motion and position-time graph at: https://brainly.com/question/2356782

Answer:

Explanation:

An atom emits a photon (particle of light) when transitioning from a ground state to its excited state. To obey conservation of energy, the energy gained by the atom when an electron moves to a lower energy level is equal to the energy it loses in emitting the photon. (The energy of a photon is E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon.) Conversely, when an atom absorbs a photon (as is the case in absorption spectra), the electron absorbing the photon moves to a higher energy level.

Answer:

not using it too much and getting addicted

Explanation:

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

Answer:

Yes

Explanation:

Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,

Hence the same amount of charge flow is delivered to any circuit.

attached below are examples

Answer:

Geothermal!!

Explanation:

Geo means earth! Just like Hydro means water, and Pyro means fire :)

Answer:

The Answer is D). Geothermal energy

Explanation:

Hope this helps!!!!!

Answer:

D

Explanation:

One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.

Answer:

h = 90.10 m

Explanation:

Given that,

A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m

The initial speed of the stone, u = 10 m/s

The path followed by the projectile is given by :

[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)

For maximum height,

Put dh/dt = 0

So,

[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]

Put the value of t in equation (1).

[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]

So, the maximum height of the stone is equal to 90.10 m.

Answer:

Maximum frequency on power spectrum plot = 101 Hz

Explanation:

Given:

Time taken for output = 15 seconds

Frequency rate = 202 Hz

Find:

Maximum frequency on power spectrum plot

Computation:

Maximum frequency = Given frequency rate / 2

Maximum frequency on power spectrum plot = Frequency rate / 2

Maximum frequency on power spectrum plot = 202 / 2

Maximum frequency on power spectrum plot = 101 Hz

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

Answer:

F' = 73.7 N

F = 8.749×10⁻³ N

a' = a =  9.83 m/s²

Explanation:

(a)

For the rock

Applying

F' = Gm'm/r²................... Equation 1

Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.

From the question,

Given: m' = 7.5 kg

Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m

Substitute these values into equation 1

F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²

F' = 7.37×10¹ N

F' = 73.7 N

Also, For the pebble,

F = GMm/r².............. Equation 2

Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth

Given: M = 8.9×10⁻⁴ kg,

Substitute into equation 2

F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²

F = 8.749×10⁻³ N

(b)

For the rock,

a' = F'/m'

Where a' = magnitude of the acceleration of the rock

a' = 73.7/7.5

a' = 9.83 m/s²

For the pebble,

a = F/M

Where a = acceleration of the pebble

a = (8.749×10⁻³)/(8.9×10⁻⁴)

a = 9.83 m/s²

Answer:

Explanation:

Given that:

the putty initial speed (u) = 9.10 m/s

distance (s) between hand and the ceiling = 3.50 m

the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.

v² - u² = 2as

Since the putty is moving in a vertical motion(i.e. in an upward direction)

v² - u² = -2gs

v² = u² - 2gs

[tex]v = \sqrt{u^2 - 2gs}[/tex]

[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]

[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]

[tex]v = \sqrt{82.81 -68.6}[/tex]

[tex]v = \sqrt{14.21}[/tex]

v = 3.77 m/s

2.

The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:

v = u + at

In an upward direction

v = u - gt

making time t the subject;

v - u = -gt

[tex]t = \dfrac{v-u}{-g}[/tex]

[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]

[tex]t = \dfrac{-5.33}{-9.8}[/tex]

t = 0.54s

Answer:

Food, clothes, language, and belief

I am sure it help you with that much ☺️

Explanation:

pleasae give me some thanks please good morning sister