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4 years ago

5

What does a lunar eclipse and a solar eclipse have in common

1 answer:

liraira [26]4 years ago

3 0

During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".

You might be interested in

A constant force of 5KN pulls a crate along a distance of 15 m in 75s.What is the power

xxMikexx [17]

Explanation:

We know,

1KN = 1000N

Then, Force(F) = 5*1000N

=5000N

Here,

Power (P)=Work(W)/Time(T)

=Force * distance/ Time (W = F*s)

= 5000*15/75

=1000

So, The power of body or object is 1000Watt.

I hope this will be helpful for you.

8 0

4 years ago

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b

Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is $n_2 = 1.333$ , and for air $n_1 = 1.00$ : the angle of light with the normal is $90^o-60^o = 30^o$ ; therefore Snell's law gives

$n_1sin(\theta_1)= n_2sin(\theta_2)$

$1.00*sin(\theta_1) = 1.33 sin(30^o)$

$sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}$

$sin (\theta_1) = 0.665$

$\theta _1 = sin^{-1}(0.665)$

$\boxed{\theta_1 = 41.68^o}$

4 0

3 years ago

Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose,

fenix001 [56]

Answer:

It will take 29.31 seconds for the Boxster to catch the Scion

Explanation:

Given the data in the question;

lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A

the distance travelled by car A is

x = $V_{A}$ × t

where $V_{A}$ is the speed of the car and t is time

the distance travelled by car B before reaching car A will be;

x + x₀ = $V_{B}$ × t

Now lets replace x by $V_{A}$ × t

so

( $V_{A}$ × t) + x₀ = $V_{B}$ × t

x₀ = ( $V_{B}$ × t) - ( $V_{A}$ × t)

x₀ = t ( $V_{B}$ - $V_{A}$ )

t = x₀ / ( $V_{B}$ - $V_{A}$ )

so we substitute

t = 170 m / (24.4 - 18.6)

t = 170 / 5.8

t = 29.31 s

Therefore; it will take 29.31 s for the Boxster to catch the Scion

8 0

3 years ago

A block is pulled across a table by a constant force of 9.20 N. If the mass of the block is 2.30kg, how fast will the block be m

kaheart [24]

Answer:

$8\:\text{m/s}$

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . Using this, we can find the acceleration of the object:

$9.20=2.30a,\\a=\frac{9.20}{2.30}=4\:\mathrm{m/s^2}$ .

Now that we've found the block's acceleration, we can use the following kinematics equation to find its final velocity after 2 seconds:

$v_f=v_i+at,\\v_f=0+4(2),\\v_f=\boxed{8\:\text{m/s}}$

*Assumption: The block is initially at rest and has a initial velocity of zero. Otherwise, the question is unsolvable.

5 0

3 years ago

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the drawstring ba

Alex

Answer:

v=39.05 m/s

Explanation:

Given that

x= 56 cm

F= 158 N

m= 58 g = 0.058 kg

Lets take spring constant = k

At the initial position,before releasing the arrow

F= k x

By putting the values

F= k x

158= 0.56 k

k=282.14 N/m

Now from energy conservation

Lets take final speed of the arrow after releasing

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2$

k x²=mv²

282.14 x 0.56² = 0.058 v²

v=39.05 m/s

3 0

3 years ago