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3 years ago

A baseball going 33.0 m/s will take what time in seconds to travel 8.1 meters?

Physics

1 answer:

Studentka2010 [4]3 years ago

3 0

Speed=33m/s
Distance=8.1m

$\\ \bull\sf\dashrightarrow Speed=\dfrac{Distance}{Time}$

$\\ \bull\sf\dashrightarrow Time=\dfrac{Distance}{Speed}$

$\\ \bull\sf\dashrightarrow Time=\dfrac{33}{8.1}$

$\\ \bull\sf\dashrightarrow Time=4.07s$

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Need Help with this physics question, just a written response

Nadusha1986 [10]

Answer:

Explanation:

According to Newton's third law, every action has an equal and opposite reaction

so it tells us that the force exerted by the earth on the spacecraft is equal to the force exerted by the spacecraft on the earth. But we do not see the earth moving towards the spacecraft because the inertia of the spacecraft is very less than the inertia of the earth.

3 0

3 years ago

A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$

Our values are:

$V=30.1m/s$

$C_d=0.45$

$A=3.3m^2$

$\rho=1.2kg/m^3$

Replacing,

$F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2$

$F_D=807.25N$

We need calculate now the energy lost through a time T, then,

$W = F_D d$

But we know that d is equal to

$d=vt$

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

$W=(807.25N)(30.1m/s)(3600s/1hr)$

$W=87.47*10^6J$ (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0

3 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

3 years ago

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

$\lambda = \frac{v}{f}$

Where,

$\lambda =$ Wavelength

f = Frequency

v = Velocity

Our values are given as,

$f = 2.8*10^3Hz$

$v = 340m/s \rightarrow$ Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

$\lambda = \frac{340}{2.8*10^3}$

$\lambda = 0.1214m$

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0

3 years ago

Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill

laila [671]

Answer:

a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d) , c) m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

the maximum for m = 1 is at the angle

θ = sin⁻¹ λ / d

the second maximum m = 2

θ = sin⁻¹ ( λ / 2d)

the third maximum m = 3

θ = sin⁻¹ ( λ / 3d)

the fourth maximum m = 4

θ = sin⁻¹ ( λ / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Ф = π d sin θ /λ

x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference d sin θ = m λ

first diffraction minimum a sin θ = λ

we divide the two expressions

d / a = m

In our case

3a / a = m

m = 3

order three is no longer visible

7 0

3 years ago