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3 years ago

TAKING 100 PONTS!!! PLEASE SOMEONE HELP.JUST LOOK THE PICTURE

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

Explanation:

oh no i hope i dont fall and the i agree with the other dude

lesantik [10]3 years ago

4 0

Answer: Okay, thanks I really appreciate the generosity.

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As outlaws escape in their getaway ear, which goes 3/4 c, the police officer fires a bullet from a pursuit ear, which only goes

Mariulka [41]

Answer:

a) Bullet will hit

b) Bullets will not hit

Explanation:

Given:

The velocity of the bullet, u = $\frac{1}{3}c$ in the rest frame of the bullet pursuit car

The velocity of the original frame of reference, v = $-\frac{1}{2}c$ with respect to the pursuit car.

Now, according to the Galileo

the velocity of the bullet in the original frame of reference (u') will be

u' = u - v

on substituting the values we get

u' = $\frac{1}{3}c-(-\frac{1}{2}c)$

u' = $\frac{1}{3}c+\frac{1}{2}c$

u' = $\frac{5}{6}c$

since this velocity ( $\frac{5}{6}c$ ) is greater than the ( $\frac{3}{4}c$ )

hence,

<u>the bullet will hit</u>

Now, according to the Einstein theory

the velocity of the bullet in the original frame of reference (u') will be

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

on substituting the values we get

$u'=\frac{\frac{1}{3}c-\frac{1}{2}c}{1-\frac{\frac{1}{3}c\times \frac{1}{2}c}{c^2}}$

$u'=\frac{\frac{5}{6}c}{1-\frac{1}{6}}$

$u'=\frac{5}{7}c$

since,

$u'=\frac{5}{7}c$ is less than ( $\frac{3}{4}c$ ), this means that the bullet will not hit

7 0

4 years ago

Develop expression for equation for uniform motion in a straight line

frozen [14]

According to cuneiform tablets in the ancient world, straight lines cannot cross, and no motion in the world is not relative. Btw...I KNOW!!! GOT MILK???

4 0

3 years ago

When does DNA replication occur?

Digiron [165]

Answer:

NA is replicated during the S phase (Synthesis phase) of Interphase.

Hope it helps!!!!!!!!!!

5 0

3 years ago

Read 2 more answers

PLEEEASE HELP ME!!! YOU GET 15 PTS

Damm [24]

Answer:

Explanation:

4 0

4 years ago

Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is kilometers, and the

serious [3.7K]

Complete question is;

Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun

Answer:

Minimum distance = 147,099,713.4 km

Maximum distance = 152,096,286.6 km

Explanation:

The formula for the eccentricity of an ellipse is given by;

e = c/a

where;

c is distance from the center of the ellipse to the focus of the ellipse.

a is distance from the center of ellipse to a vertex.

From the question, we are given;

a = 149,598,000

e = 0.0167

Thus;

0.0167 = c/149,598,000

c = 0.0167 × 149,598,000

c = 2,498,286.6

Now, formula for the minimum distance (perihelion) is;

Minimum distance = a - c

Minimum distance = 149598000 - 2498286.6

Thus;

Minimum distance = 147,099,713.4 km

Similarly, formula for the maximum distance (aphelion) is;

Max distance = a + c

Max distance = 149598000 + 2498286.6

Maximum distance = 152,096,286.6 km

4 0

3 years ago

TAKING 100 PONTS!!! PLEASE SOMEONE HELP.JUST LOOK THE PICTURE​

TAKING 100 PONTS!!! PLEASE SOMEONE HELP.JUST LOOK THE PICTURE