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3 years ago

if a car has a constant acceleration of 4 m/s^2, starting from rest, how fast is it traveling after 5 seconds?

Physics

2 answers:

UkoKoshka [18]3 years ago

3 0

Answer:

20 m/s

Explanation:

Recall that one of the equations of motions can be written:

v = u + at, (also see attached for reference)

Where,

v = final speed (we are asked to find this)

u = initial speed = 0 (because it starts from rest)

t = time taken = 5s

We simply substitute the given values into the equation:

v = u + at

v = 0 + (4)(5)

v = 20 m/s

Anni [7]3 years ago

3 0

Answer:If a car has a constant acceleration of 4 m/s2, starting from rest, how far has it traveled after 5 seconds? The distance travelled in 5s.

Explanation:

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A wooden rod of negligible mass and length 84.0 cm is pivoted about a horizontal axis through its center. A white rat with mass

Mandarinka [93]

Answer:

v = 2.029 m/s

Explanation:

Given

L = 84.0 cm ⇒ R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m

m₁ = 0.600 kg

m₂ = 0.200 kg

g = 9.8 m/s²

u₁ = u₂ = 0 m/s

v₁ = ?

v₂ = ?

Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).

In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.

then

Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J

Ef = Kf + Uf

⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²

⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464

⇒ Ef = Kf + Uf = 0.4*v² - 1.6464

Since

0 = 0.4*v² - 1.6464 ⇒ v = 2.029 m/s

v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.

v₁ = v₂ = v ⇒ ω₁*R₁ = ω₂*R₂ ⇒ ω₁*R = ω₂*R ⇒ ω₁ = ω₂ = ω

⇒ v = ω*R

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3 years ago

A light-year is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many miles are there in o

Alinara [238K]

Answer:

(a) $5.88\times 10^{12}mi$

Explanation:

We have given speed of light $c=3\times 10^8m/sec$

Given time = 1 year =365 days

We know that in 1 minute = 60 sec

1 hour = 60×60 = 3600 sec

In one day = 24 hour = 24×60×60=86400 sec

So in 365 days = 365×86400 $=3.1536\times 10^7sec$

We know that distance = speed ×time $=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m$

We have given that 1 mi = 1609 m

So $9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi$ So option (a) is correct

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?

Xelga [282]

<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

30 cm = 0.3 m (distance)

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

$\small \boxed{work \: done \: by \: Ryan \: = 3 \: Joules}$

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2 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

2 years ago

if a car has a constant acceleration of 4 m/s^2, starting from rest, how fast is it traveling after 5 seconds?​

if a car has a constant acceleration of 4 m/s^2, starting from rest, how fast is it traveling after 5 seconds?