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3 years ago

14

What is the highest point on a wave called

1 answer:

kirill [66]3 years ago

8 0

It can be called the 'crest' or the 'peak' of the wave.

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On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on

Lelu [443]

Answer:

True.

Explanation:

He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater. Hence you will be drown inside the water.

8 0

2 years ago

Please help ! Which of the following objects has the greatest momentum?

shusha [124]

Answer:

maybe the third one....

8 0

2 years ago

A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The

nadya68 [22]

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

$\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s$

So the change in momentum of the ball would be the product of its velocity change and its mass

$\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s$

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

8 0

2 years ago

A 42.6kg lamp is hanging from wires as shown in figure.The ring has negligible mass. Find tensionsT1, T2,T3 if the object is in

IRISSAK [1]

Answer:

T1 = 417.48N

T2 = 361.54N

T3 = 208.74N

Explanation:

Using the sin rule to fine the tension in the strings;

Given

amass = 42.6kg

Weight = 42.6 * 9.8 = 417.48N

The third angle will be 180-(60+30)= 90 degrees

Using the sine rule

W/Sin 90 = T3/sin 30 = T2/sin 60

Get T3;

W/Sin 90 = T3/sin 30

417.48/1 = T3/sin30

T3 = 417.48sin30

T3 = 417.48(0.5)

T3 = 208.74N

Also;

W/sin90 = T2/sin 60

417.48/1 = T2/sin60

T2 = 417.48sin60

T2 = 417.48(0.8660)

T2 = 361.54N

The Tension T1 = Weight of the object = 417.48N

8 0

2 years ago

A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a

dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

$A_1v_1 = 18 A_2v_2$

now we know that

$A_1 = 1 cm^2$

$A_2 = 0.4 cm^2$

now from above equation

$1 cm^2 v_1 = 18(0.400 cm^2)v_2$

$\frac{v_2}{v_1} = \frac{1}{18\times 0.4}$

$\frac{v_2}{v_1} = 0.14$

so velocity will reduce by factor 0.14

3 0

3 years ago