Ask question

Ask question

All categories

2 years ago

13

If you are driving 95 km/h along a straight road and you look to the side for 2.0s, how far do you travel during this inattentiv

e period

1 answer:

Bad White [126]2 years ago

3 0

95km/h
95km/60*60sec
(95/(60*30))km/2sec
0.05277km

You might be interested in

A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o

vredina [299]

Answer:

Recoil speed, $1.17\times 10^{-5}\ m/s$

Explanation:

Given that,

Mass of the comet fragment, $m_1=1.96\times 10^{13}\ kg$

Speed of the comet fragment, $v_1=6.5\times 10^4\ m/s$

Mass of Callisto, $m_2=1.08\times 10^{23}\ kg$

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

$m_1v_1=(m_2+m_2)V$

V is recoil speed of Callisto immediately after the collision.

$V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s$

So, the recoil speed of Callisto immediately after the collision is $1.17\times 10^{-5}\ m/s$

7 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0

2 years ago

What is a volcanically active area that is not near a tectonic plate boundary?

belka [17]

It a blue tectonic movement

8 0

2 years ago

If u have 2500Watts,240Volts . its resistance

Virty [35]

Answer:

2500/240 will give the resistance to be 10.42

3 0

2 years ago