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3 years ago

15

A car is traveling at 16 m/s^2

1 answer:

Leya [2.2K]3 years ago

4 0

Really? wow that's pretty cool

You might be interested in

Newtown third law applies to blank of objects

fomenos

Answer:

All

Explanation:

I'm not sure what you meant but Newton's third law which basically states that every action has an equal and opposite reaction applies to <em>all</em> objects. So I think the answer is all.

8 0

3 years ago

The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m

IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

$\rho$ = Resistivity of gold = $2.44\times 10^{-8}\ \Omega .m$ (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = $\dfrac{\pi}{4}d^2$

E = Electric field

Resistivity is given by

$\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m$

The electric field in the wire is 0.0360531138247 V/m

6 0

3 years ago

Read 2 more answers

Assuming that the hill on the left will provide all of the potential energy for the ride, would this roller coaster actually mak

trapecia [35]

Yes I believe it would I think the answer is D

3 0

3 years ago

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$

Explanation:

Given

$I_1=2 Amps$

$I_2=0.75 Amps$

Length of each wires $\left ( L\right )=5 m$

Distance between wires $\left ( r\right )=5 cm$

Force per unit length = $\frac{\mu_0I_1I_2}{2\pi r}$

$F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}$

$F'=6\times 10^{-6}$

Force for $L=5 m$

$F=3\times 10^{-5} N$

6 0

3 years ago

When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr

Scrat [10]

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

7 0

3 years ago