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2 years ago

A porsche 911 ca go from 0 mph to 60 mph in 3.4secs whats ur acceleration

Physics

1 answer:

lawyer [7]2 years ago

8 0

The acceleration is $7.9 m/s^2$

Explanation:

The acceleration of a body is given by

$a=\frac{v-u}{t}$

where

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we first have to convert the velocities into SI units. Keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We have:

$u=0 mph = 0 m/s$

$v=60 mph \cdot \frac{1609 m/mi}{3600 s/h}=26.8 m/s$

And the time interval is

t = 3.4 s

Therefore the acceleration is

$a=\frac{26.8-0}{3.4}=7.9 m/s^2$

Learn more about acceleration:

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brainly.com/question/11181826

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

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2 years ago

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

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2 years ago

The vocal tract is a resonator, and the transmission of a sound through an acoustic resonator is highly dependent upon frequency

andriy [413]

Answer:

Your question is not complete. this option completes your question

Which of the following is not a formant wavelength?

A. 3λ/4.

B.λ /4.

c. λ/2

D. 7λ/4

Explanation:

A formant wavelength will be such that the total length of pipe = nλ/2 + λ/4.

When the above condition is satisfied then only a node is formed at the closed end and antinode at the open end.

Among the given options:

(a) 7λ/4 = nλ/2 + λ/4, satisfied for n = 3

(b) 3λ/4 = nλ/2 + λ/4, satisfied for n = 1

(d) λ/4 = nλ/2 + λ/4, satisfied for n = 0

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2 years ago

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m

Anit [1.1K]

(a)

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.

(b)
To much the same end do we derive the vertical component:

18.0*sin37.5 = v_y = 10.96 ms^-1

Which we then divide by acceleration, a_y, to derive the time till maximal displacement,

10.96/9.8 = 1.12 s

Finally, doubling this value should yield the particle's total time with r_y > 0

2.24 s

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

6 0

2 years ago