What color is absorbed when we see a yellow flower?

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

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

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

2 years ago

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

2 years ago

High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g

tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, $p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s$

After the collision, final momentum $p_f=0.21\ kg\times 42\ m/s+0.046v$

Using the conservation of momentum as :

$p_i=p_f$

$11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v$

v = 63.91 m/s

So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0

2 years ago

26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the

Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

= 4N/200N

= 0.02.

7 0

2 years ago

In order for a current to be maintained through a conductor a difference in

algol13

Answer:

Potential

Explanation:

Potential difference is otherwise known as voltage. With no voltage flowing through a conductor a current will not flow.

5 0

2 years ago