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ahrayia [7]
3 years ago
11

An unfortunate bug splatters against the windshield of a moving car. Compared to the force of the car on the bug, the force of t

he bug on the car is
Physics
1 answer:
irina [24]3 years ago
7 0

"the same"


Hope I helped

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Good morning friends​
Hatshy [7]

Answer:

Good morning. Whats about you

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3 years ago
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A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made
olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

B=\dfrac{\mu_oI}{2\pi r}

B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}

B = 0.000002 T

B=10^{-5}\ T

(2) Exact formula:

B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }

B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

4 0
3 years ago
What is the momentum of an 18-kg object moving at 0.1 m/s ?
Anit [1.1K]

Answer:

1.8 m/s

Explanation:

here's the solution : -

momentum = mass × velocity

=》18 × 0.1

=》1.8 m/s

6 0
3 years ago
One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a
Novay_Z [31]

Answer:

1.60 g

Explanation:

From the attached file below:

we can deduce that:

v =  v_x =v_y = 20 \ m/s \\t = 2s

The distance traveled by 2 s will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The length is quarter of the circle  with radius r,

so; if 2 πr = 4 x

Then radius (r) will be:

r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}

r = 25.5 m

The centripetal acceleration can be expressed as:

a = \frac{v^2}{r}

so;

a = \frac{(20 \ m/s^2)}{25.5 \ m}

a = 15.7 m/s²

the magnitude of the acceleration experienced by your unfortunate passengers in terms of acceleration due to gravity is then determined by the equation:

a' = \frac{a}{g} g

a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g

a' = 1.60 \ g

∴ The magnitude of the acceleration experienced by your unfortunate passengers during the turn = 1.60 g

6 0
3 years ago
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Which pm the following are likely to form a covalent bond
Alecsey [184]

Answer:

magneisuim and gold

Explanation:

8 0
4 years ago
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