In machines with low efficiency, most of the lost energy is turned into?

a. Light
b. Sound
c. Power
d. Heat

Answer:

Explanation:

Physics is the branch of science that studies the natural world and its rules and orders. It is one of the oldest sciences as ancient people study the stars and astronomy is considered part of Physics.

To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

Answer:

Explanation:

this is like rubbing a balloon on your head to make your hair stand up.  Do that to the can.    The balloon is filled , ofc,  and then just rub the balloon on the can.  This will charge the can with static electricity.   :P

Answer:

E

Explanation:

Planets after Mars in our solar system are called Jovian planets. Therefore, Jupiter, Saturn, Uranus and Neptune are Jovian planets. The specialty of these planets is that they mostly made of gases and have ring around them.

They have rings around them because tidal forces cause volcanic eruptions on some moons, and part of this material subsequently escaped the gravity of the moons, forming the rings.

The index of refraction is 1.33.

To find the answer, we need to know about index of refraction.

What is index of refraction?

What is the mathematical expression of refractive index?

n= sin∅₁ / sin∅₂

where, ∅₁= angle of incident

             ∅₂= angle of refraction

What is the refractive index,  if the angle of incidence in air is 40 degrees and the angle of refraction is 29 degrees?

Thus, we can conclude that the index of refraction is 1.33.

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Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

          a sin θ = m λ

         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

        sin θ  = λ / a

angles are generally very small and are measured in radians

         sin θ  = θ  = y / x

we substitute

         [tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]

the width of the central maximum is twice the distance to zero

         w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

          W = 2x

           2y = 2x

we substitute

          1 = λ/ a

          λ = a

we see that the width of the slit is equal to the wavelength used.

Answer:

8.33 meters/sec.

time = 30 sec. 30 sec. = 8.33 meters/sec.

Answer:

d is trueeeeeeeeeeeeeeeee

Answer:

The neutral atom becomes an anion.

Explanation:

When a neutral atom gains an electron (e−), the number of protons (p+) in the nucleus remains the same, resulting in the atom becoming an anion (an ion with a net negative charge).

Answer:

5.4 will be the weight in illustrate form

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex] (1)

Where:

[tex]I_{T}[/tex] - Moment of inertia of the turntable, in kilogram-square meters.

[tex]r[/tex] - Distance of the block regarding the center of the turntable, in meters.

[tex]m[/tex] - Mass of the object, in kilograms.

[tex]\omega_{o}[/tex] - Initial angular speed of the turntable, in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex] (2)

Where [tex]M[/tex] is the mass of the turntable, in kilograms.

If we know that [tex]\omega_{o} \approx 7.330\,\frac{rad}{s}[/tex], [tex]M = 1.5\,kg[/tex], [tex]m = 0.54\,kg[/tex] and [tex]r = 0.1\,m[/tex], then the angular speed of the turntable after the event is:

[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex]

[tex]I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}[/tex]

[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex]

[tex]\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}[/tex]

[tex]\omega_{T} = 3.004\,\frac{rad}{s}[/tex] ([tex]28.687\,\frac{rev}{min}[/tex])

The turntable's angular speed after the event is 28.687 revolutions per minute.

Answer:

[tex]\frac{T_t}{T_c} = 1.32[/tex]

Explanation:

The torque applied on an object can be calculated by the following formula:

[tex]T = Fr[/tex]

where,

T = Torque

F = Applied Force

r = radius of the wheel

For car wheel:

[tex]T_c = Fr_c\\[/tex]

For truck wheel:

[tex]T_t = Fr_t[/tex]

Dividing both:

[tex]\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}[/tex]

for the same force applied on both wheels:

[tex]\frac{T_t}{T_c} = \frac{r_t}{r_c} \\[/tex]

where,

rt = radius of the truck steering wheel = 0.25 m

rc = radius of the car steering wheel = 0.19 m

Therefore,

[tex]\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\[/tex]

[tex]\frac{T_t}{T_c} = 1.32[/tex]

Answer:

Subduction , Latin for "carried under," is a term used for a specific type of plate interaction. It happens when one lithospheric plate meets another—that is, in convergent zones —and the denser plate sinks down into the mantle.

Answer:

[tex]0.3\ \text{T}[/tex]

Explanation:

q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]

m = Mass of particle = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]

v = Velocity = [tex]2\times 10^5\ \text{m/s}[/tex]

E = Electric field = [tex]6\times 10^4\ \text{V/m}[/tex]

B = Magnetic field

Magnetic field is given by

[tex]B=\dfrac{E}{v}\\\Rightarrow B=\dfrac{6\times 10^4}{2\times 10^5}\\\Rightarrow B=0.3\ \text{T}[/tex]

The magnitude of magnetic field is [tex]0.3\ \text{T}[/tex]

Answer:

I=R×V

360×120 V

=43,200 A

Answer:

A.)0.33 A

Explanation:

i just took the quiz 2021

Answer:

pyramid, cone, sphere, cube

Answer:

the frequency range of sound waves is about 70 Hz - 160 Hz

Hence, Option a)  About 70 Hz -160 Hz is the correct answer

Explanation:

Given the data in the question;

L₁ = 10 micrometers = 0.00001 m

so

T₁ = 2π√(L/g)

g = 9.8 m/s

so we substitute

T₁ = 2π√(0.00001 /9.8) = 0.006347

⇒ f₁ = 1 / T₁  = 1 / 0.006347 = 157.55 ≈ 160 Hz

L₂ = 50 micrometers = 0.00005 m

T₂ = 2π√(L/g)

g = 9.8 m/s

so we substitute

T₂ = 2π√(0.00005 /9.8) = 0.0142

f₂ = 1 / T₂ = 1 / 0.0142 = 70.422 ≈ 70 Hz

Therefore, the frequency range of sound waves is about 70 Hz - 160 Hz

Hence, Option a)  About 70 Hz -160 Hz is the correct answer

Answer:

w = w₀ / 2    the angular velocity is half the initial value.

Explanation:

We can analyze this exercise as if we added another disk to obtain a disk with twice the mass, for which if the system is two disks, the angular tidal wave is conserved

initial instant.

          L₀ = I₀ w₀

final moment

          L_f = I w

the moment is preserved

          L₀ = L_f

          I₀ w₀ = I w

the moment of inertia of a disk is

         I = ½ m R²

we substitute

          ½ m R² w₀ = ½ (2m) R² w

          w = w₀ / 2

for the case of a disk with twice the mass, the angular velocity is half the initial value.

Answer:

[tex]4.25\ \text{m/s}[/tex]

[tex]3391.22\ \text{N}[/tex]

Explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Energy balance of the system is given by

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]

The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]

[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]

The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].

The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.

An equation is an expression that shows the relationship between two or more numbers and variables.

The two points on a progressive wave are out of phase by 0.41 rad.

Hence, Phase difference = 0.41 rad

But:

Rad to degree = (rad * 180/π)°

Hence:

0.41 rad = (0.41 rad * 180/π) = 23°

The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.

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Answer:

Yes, it would make it back up.

Explanation:

If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.

I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.

The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.

Answer:

the height of the stadium is 4 m

Explanation:

The computation of the height of the stadium is shown below:

but before that total mechanic energy should be determined

E = PE + KE

where

PE = mgh

and, KE = 1 ÷2 mv^2

Now

E = mgh + 1 ÷2 mv^2

= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2

= 71.9J

= 72J

Now the height of the stadium is

TE = mgh

72 = (1.8) × (9.8) × h

So, h = 4 m

Hence, the height of the stadium is 4 m

Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

A)    t = 10.56 s, B)  x = 235 m, C) v = 25.2 m / s

Explanation:

A) We can solve this problem using kinematics expressions.

The distance traveled by the truck is

       x_c = v_c t

Distance traveled by the car.

The car must travel the distance that separates them from the truck x₀=25.0.   Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2  4.5 = 9 m, therefore the total length traveled by the car is

          x_t = x₁ + x₂ + x₃

          x_t = 26.5 + 20.7 +9 = 56.2 m

the distance traveled by the car when it returns to the lane is

         x_c + x_t = x₀ + v₀ t + ½ a t²

when the car passes the car the distance traveled by the two vehicles is the same, we substitute

         v_c  t + x_t = x₀ + v₀ t + ½ a t²

         ½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0

we substitute the values

         ½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =

         0.28 t² -31.2 = 0

         t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]

         t = 10.56 s

This is the time it takes for the car to pass the truck and back into the lane.

B) the distance traveled is

        x = v₀ t + ½ a t²

        x = 19.3 10.56 + ½ 0.560 10.56²

        x = 235 m

C) the final velocity is

         v = v₀ + a t

         v = 19.3 + 0.560 10.56

         v = 25.2 m / s

Answer:

Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.