it takes a cheetah just 3s to reach its top speed of 30m/s what is it's acceleration​

Answer:

"Cannot be determined," if that's an answer choice. It depends on the velocities of both objects, since momentum=mass*velocity.

Explanation:

Answer:

it is long.

Explanation:

I know the answer but I don't know how to explain it

Answer:

[tex]23796.19\ \text{N}[/tex]

[tex]0.0002032\ \text{s}[/tex]

Explanation:

F = Force

s = Displacement = 6.3 cm

m = Mass of bullet = 7.8 g

v = Velocity of bullet = 620 m/s

t = Time taken

Work done is given by

[tex]W=Fs[/tex]

Kinetic energy is given by

[tex]K=\dfrac{1}{2}mv^2[/tex]

Using work energy considerations we get

[tex]Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}[/tex]

The average force that stops the bullet is [tex]23796.19\ \text{N}[/tex].

Force is given by

[tex]F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}[/tex]

The time taken to stop the bullet is [tex]0.0002032\ \text{s}[/tex].

Answer:

the fundamental frequency produced by the open pipe is 120.78 Hz

Explanation:

Given;

length of the open pipe, L = 1.42 m

speed of sound in air, v = 343 m/s

The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

The fundamental frequency produced by the open pipe is calculated as;

[tex]f_o = \frac{v}{\lambda} \\\\f_o = \frac{v}{2L} \\\\f_o = \frac{343}{2 \times 1.42} \\\\f_o = 120.78 \ Hz[/tex]

Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz

Answer:

true for first and false for second

Explanation:

Answer:

Potential Energy

Explanation:

Answer:

a) the weight of the person is 170.5 lbf

b) weight of the astronaut on the moon is 28.2 lbf

Explanation:

Given the data in the question;

a)

we know that;

weight on the surface of the earth = m[tex]g_{earth[/tex]

given that m = 170.5 lbm and g = 32.174 ft/s²

we substitute

weight on the surface of the earth =  170.5 lbm × 32.174 ft/s²

= 5485.667 lbm-ft/s²

1 lbf = 32.174 lbm-ft/s²

so

weight on the surface of the earth = (5485.667 / 32.174) lbf

weight on the surface of the earth = 170.5 lbf

Therefore, the weight of the person is 170.5 lbf

b)

given that;

weight on the surface of the earth = m[tex]g_{moon[/tex]

m = 170.5 lbm and g = 5.32 ft/s²

weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²

= 907.06 lbm-ft/s²

1 lbf = 32.174 lbm.ft/s²

weight on the surface of the earth = ( 907.06 / 32.174 ) lbf

weight on the surface of the earth = 28.2 lbf

Therefore, weight of the astronaut on the moon is 28.2 lbf

Answer:

a) False, b) True, c) true, d) true, e) True, g)  False

Explanation:

Se pide que analicemos cada expresión, f) true

a) False.   La  antena vertical tiene cargas eléctricas que se pueden mover en la direccion vertical, una onda electromagnética polarizada horizontal el campo eléctrico esta en la dirección horizontal por lo cual no puede mover los electrones que están en direccional vertical

be) True.  La radiación ultravioleta es absorvidad por la melanina del la piel y crea las quemaduras comunes de insolación.

C) False.  Todas la ondas electromagnetismo viajan en la misma velocidad c

d) True.  La Temperatura del sol es 5800K por lo cual la radiación mas intensa esta en el rango visible

e) True.  Los rayos gamma son de muy alta energía, por lo cual se producen en ransicio0n de los núcleos atmicos.

f) True.  El rango de frecuencia para las onda de radio va desde 500 Kahoy hasta 110 MHz, por lo tanto el rango de longitudes de onda es

           c= lam f

           lam = c/f

           lam = 3 108/500 103

           lam = 6 102 m

           lam = 3 108 / 110 106  

           lam = 2,72 m

los lo ttnato en todo el rango de las frecuecias de radio la lambda> 1 m

g) False.  Todas las ondas electromagnéticas viajan a la misma velocidad c( 3 108m/s)

Answer:

The correct answer is - 1240 newtons; 278.763 lbs.

Explanation:

Answer:

Six purposes of government: to form a more perfect union; establish justice; insure domestic tranquility; provide for the common defense; promote the general welfare; secure the blessings of liberty now and in the future.

Explanation:

NASA's Mars Pathfinder mission, launched in 1996, put the first free-moving rover—called Sojourner—on the planet. ... Together, these missions have shown scientists that Mars is an active planet that is rich in the ingredients needed for life as we know it—water, organic carbon, and an energy source

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load [tex]m_{LS}[/tex] = 10000 kg

mass of flat bed [tex]m_{FB}[/tex] = 20000 kg

initial speed of truck [tex]v_{0}[/tex] = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs [tex]F_{N}[/tex]     -------------let this be equation 1

where [tex]F_{N}[/tex] = normal force = mg

so

Fs,max = μs mg

ma[tex]_{max}[/tex] = μs mg

divide through by mass

a[tex]_{max}[/tex] = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²

a[tex]_{max}[/tex] = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx

where [tex]v_{f}[/tex] is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Answer:

it should be tension for the space

Answer:

tension

Explanation:

I'm not sure if I'm right but if the example is a pulley a pulley applies tension to a belt or rope or chain to keep it tight

Answer:

30°

Explanation:

The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.

∠i = ∠r

so as ∠i = 30°

∠r will also be 30°

Answer:

The amplitude decreases

Explanation:

From the inverse square law, we know that;

I ∝ 1/d²

Where;

I is intensity

d is distance

Also, we know that;

A² ∝ I

Where A is amplitude

Thus, we can say that;

A ∝ 1/d²

Thus, amplitude is inversely proportional to the distance.

So the larger the distance, the lesser the amplitude and the lesser the distance, the higher the amplitude.

Thus, as you move farther from the speaker the distance increases and therefore the amplitude decreases

The IDEAL mechanical advantage of this ramp is (25m / 5m) = 5 .

But it's only giving you a real MA of (4500N/1000N) = 4.5 .

The friction between the crate and the surface of the ramp is robbing some of the work you do as you slide the crate up the ramp, which degrades the mechanical advantage.

Answer:

The sun and the stars

Explanation:

I hope this helps!

Answer:

Northstar & North Pole

Explanation:

yyggggggggggggggg

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

Answer:

momentum = mass * velocity ... kg•m/s

acceleration = v / t = 200 m/s / 11 ms ... m/s^2

force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N

Explanation:

Answer:

Distillation separates a liquid from a solution. For example, water can be separated from salty water by simple distillation. This method works because the water evaporates from the solution, but is then cooled and condensed into a separate container. The salt does not evaporate and so it stays behind.

Answer:

25000J

Explanation:

Formula : Q = m×c×Δt

Q=Heat energy

m= mass

c=specific heat capacity

ΔT = change in temperature.

Q=M x C for this question.

Specific heat capacity = 2500 J/Kg

Mass = 10kg

2500 x 10 = 25000 J

Answer:

checking your e-mail on your computer using a coffee's shop WiFi. No need radio waves for this.

Explanation:

Because you are only checking it, nothing else.

Answer:

[tex]31.035^{\circ}[/tex]

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

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

Displacement in x direction is given by

[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]

Displacement in y direction is given by

[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]

[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]

The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].

Answer:

[tex]3.031\times 10^{-5}\ \text{m}[/tex]

Explanation:

[tex]y[/tex] = Distance between central maxima and first minimum

m = Order = 1

d = Thickness of hair

[tex]\lambda[/tex] = Wavelength = 632.8 nm

L = Distance between light source and screen = 1.25 m

Width of central maximum is given by

[tex]2y=5.22\times 10^{-2}\\\Rightarrow y=\dfrac{5.22\times 10^{-2}}{2}\\\Rightarrow y=0.0261\ \text{m}[/tex]

Distance between central maxima and first minimum is given by

[tex]y=L\tan\theta_{min}\\\Rightarrow \tan\theta_{min}=\dfrac{y}{L}\\\Rightarrow \tan\theta_{min}=\dfrac{0.0261}{1.25}\\\Rightarrow \theta_{min}=\tan^{-1}0.02088\\\Rightarrow \theta_{min}=1.1962^{\circ}[/tex]

Since [tex]\theta[/tex] is small [tex]\tan\theta_{min}=\sin\theta_{min}[/tex]

[tex]\sin\theta_{min}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1\times 632.8\times 10^{-9}}{\sin1.1962^{\circ}}\\\Rightarrow d=3.031\times 10^{-5}\ \text{m}[/tex]

The strand of hair is [tex]3.031\times 10^{-5}\ \text{m}[/tex] thick.

Answer:

The correct option is b: 30 N.

Explanation:

First, we need to find the acceleration due to gravity (a):

[tex] y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} [/tex]   (1)

Where:

[tex]y_{f}[/tex]: is the final vertical position (obtained from the graph)

[tex]y_{0}[/tex]: is the initial vertical position (obtained from the graph)

v₀: is the initial speed = 0 (it is released from rest)

Δt: is the variation of time (from the graph)

From the graph, we can take the following values of height and time:

t₀ = 0 s → [tex]t_{f}[/tex] = 5 s

y₀ = 300 m → [tex]y_{f}[/tex] = 225 m

Now, by entering the above values into equation (1) and solving for "a" we have:

[tex] a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} [/tex]

Finally, the weight of the object is:

[tex] W = ma = 5 kg*6 m/s^{2} = 30 N [/tex]

Therefore, the correct option is b: 30 N.

I hope it helps you!                                                                                                            

Answer:

C. Plate Tectonics

Explanation:

The theory of plate tectonics is when the lithosphere is separated into plates. These plates move over or float over the asthenosphere. The movement of these plates cause earthquakes and can interact with the volcanic activity.