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VashaNatasha [74]

3 years ago

15

A friend tells you that a rowboat is propelled forward by the force of its oars against the water. First explain whether the sta

tement is correct and then identify the action and reaction forces

1 answer:

stellarik [79]3 years ago

7 0

It is correct, the action is paddling, where you move the water backwards, and the reaction is the boat moving forwards.

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The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength $\lambda =1.4cm =0.014m$

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by $D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m$

So distance D will be 49.34 m

4 0

3 years ago

A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding

sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

5 0

4 years ago

For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

3 years ago

You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbita

meriva

Answer:

11.9 years

Explanation:

We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:

$\frac{T^2}{r^3}=const.$

Using the Earth as reference, we can re-write the law as

$\frac{T_e^2}{r_e^2}=\frac{T_j^2}{r_j^3}$

where

Te = 1 year is the orbital period of the Earth

re = 1 AU is the average distance of the Earth from the Sun

Tj = ? is the orbital period of Jupiter

rj = 5.20 AU is the average distance of Jupiter from the Sun

Substituting the numbers and re-arranging the equation, we find:

$T_j=\sqrt{\frac{T_e^2 r_j^3}{T_j^2}}=\sqrt{\frac{(1 y)^2 (5.2 AU)^3}{(1 AU)^3}}=11.9 y$

4 0

3 years ago

Read 2 more answers

HELP PLEASE<br>I need help on these 2 questions

Olegator [25]

I will be brief

5. There is no displacement as it didn't move

6. Time cannot go backwards so it cannot be negative

8 0

4 years ago