which of the following cannot be increased by using a machine of some kind? work, force, speed, torque

Complete Question

The Complete Question is Attached below

Answer:

[tex]E_t=2.14*10^{-13}J[/tex]

Explanation:

From the question we are told that:

Kinetic energy [tex]K.E=5*10^{-14}[/tex]

Generally, the equation for Total Energy is mathematically given by

[tex]E_t=K.E+2me[/tex]

Where

Mass of an electron in MeV

[tex]m=0.510 998 950 00 MeV[/tex]

And

Electron Charge

[tex]1.6 * 10^{-19}[/tex]

Therefore

[tex]E_t=K.E+2me[/tex]

[tex]E_t=5*10^{-14}*0.510 998 950 00*1.6 * 10^{-19}[/tex]

[tex]E_t=2.14*10^{-13}J[/tex]

dharna is said to be concentration.. it is true

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

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F = m • a

F = (15 kg) • (4 m/s^2)

F = 60 Newtons

NOTE:

That's NOT a "resultant force".

That's the force you have to EXERT on 15 kg to MAKE it accelerate at 4 m/s^2 .

Using Newton second law

[tex]\\ \sf \longmapsto Force=Mass\times Acceleration[/tex]

[tex]\\ \sf \longmapsto Force=15(4)[/tex]

[tex]\\ \sf \longmapsto Force=60N[/tex]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Answer:

convex mirror

.....................

Answer:

convex mirror..........

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step by step

formular v^2 = u^2+2as

stop v = 0

0 = 400+2a(20)

-400=40a

a = -10 m/s^2

ans affect acceleration is 10 m/s^2

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.

Answer:

Answer:

true

Explanation:

forces require an agent

you should always be able to identify what (the agent) is producing the force

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

a ball is thrown vertically upward from the ground with the velocity of 30m/s. a) how long will it take to rise to its highest poitn? b) how high does the ball rise? c) how long after projection will the ball have a velocity of 10m/s? d) what is the total time of flight?

From Vf = Vo - gt, 0 = 30 - 9.8t yielding t = 3.06 sec.

From h = Vo(t) - g(t^2)/2, h = 30(3.06) - 4.9(3.06)^2 yielding h = 45.88m.

From Vn = Vo - gt, 20 - 9.8t yielding t = 2.04 sec.

Since the fall time equals the rise time, the total flight time is 2(3.06) = 6.12 sec.

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

Answer:

wegkwe fhkrbhefdb

Explanation:B

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

Complete Question

(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?

Answer:

[tex]Q=4.50 *10^7J[/tex]

Explanation:

From the question we are told that:

Time [tex]t=5hours[/tex]

Temperature rise [tex]dT= 65\textdegree[/tex]

Power [tex]P=2500.0 Watts[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=\frac{Q}{t}[/tex]

Therefore

[tex]Q=2500*5*360[/tex]

[tex]Q=4.50 *10^7J[/tex]

Answer:

well its about our thinking but i do believe in ghost a little

Answer:

Change in state from liquid to solid.

Explanation:

Decrease in thermal energy will decrease energy and particles will slow down. Change in state from liquid to solid.

The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.

MA= Effort Distance / Load Distance

Explanation:

The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by

[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]

[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]

[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]

The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by

[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]

[tex]\:\:\:\:= 26.1°[/tex]

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

The answer is 40 kg. m/s.

Formula for momentum:

p=mv

p=(10 kg.)(4 m/s)

So, therefore, the final answer is p=40 kg. m/s.

I hope this helped answer your question. Enjoy your day, and take care!

Answer: its 400 n/s

Explanation:

cus my thing said it was right