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AlexFokin [52]
3 years ago
9

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.

Physics
2 answers:
suter [353]3 years ago
8 0

Hello!

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s² what is the cars speed after it has traveled 200 m.

We have the following data:

Vf (final velocity) = ? (in m/s)

Vi (initial velocity) = 0 m/s (accelerated from rest)

d (displacement) = 200 m

a (acceleration) = 3.90 m/s²

Since we do not need to know the time elapsed during the movement, we apply the data of the question to the Equation of Torricelli, let us see:

V_f^2 = V_i^2 + 2*a*d

V_f^2 = 0^2 + 2*3.90*200

V_f^2 = 0 + 1560

V_f^2 = 1560

V_f = \sqrt{1560}

\boxed{\boxed{V_f \approx 39.5\:m/s}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

snow_tiger [21]3 years ago
6 0
The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = \sqrt{400/3.9} \approx 10.12739367, at = v, so 3.9 * 10.12739367 \approx 39.5  m/s
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The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

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3 years ago
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Answer:

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Explanation:

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Read 2 more answers
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
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