If an automobile airbag inflates in 25 µs, what is the time in seconds?

Taking test, need physics help<br> ASAP please!

Anuta_ua [19.1K]

Answer:

im pretty sure its c the third answer i got that one right

Explanation:

your welcome :)

8 0

2 years ago

FORMULA PLEASE AND explINATION

Alex_Xolod [135]

2 Newtons to the right.

3 newtons are needed to over come the friction. There are 2 left over.

So the answer is 2 newtons to the right.

5 - 3 = 2

7 0

3 years ago

A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?

RUDIKE [14]

Answer:

Explanation:

Step one:

given data

mass of bullfrog= 0.54kg

Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

F=0.54*9.81

F=5.3N

The normal force is 5.3N

7 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

Study the diagram. Point C identifies the _____ of the wave.

Sonbull [250]

Answer:

Amplitude : The height of the wave from the origin to the crest/peak or trough

Explanation:

5 0

3 years ago

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