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3 years ago

14

a car company claims that its car can accelerate from rest to a speed of 28.0 m/s in 20.0 s. find the average acceleration of th

e car

1 answer:

elixir [45]3 years ago

5 0

tje average acceleration of the car is

$\frac{28}{20} = 1.4$

the average acceleration of the car is 1.4 m/s

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Scientists could investigate this idea by making cathode ray tubes out of different materials to see if the ray was the same.

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3 years ago

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Write a general scientific question that you will answer by doing this experiment.

il63 [147K]

Answer:

How does newtons first two laws of motion apply to the toy car?

Explanation:

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3 years ago

Desperate for help! please!!!!!!!!

lys-0071 [83]

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

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3 years ago

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Luba_88 [7]

Answer:

C

Explanation:

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3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

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3 years ago