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3 years ago

7

The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?

1 answer:

swat323 years ago

4 0

C is the answer

The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?

a.

It faces the moon more often than other places on Earth.

b.

It has many rocky beaches.

c.

It is a long, narrow inlet.

d.

Its tides cannot be predicted accurately.

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When the blood flows through the capillary bed, most of the plasma will return to the heart through the venules. What happens to

Bas_tet [7]

The answer is the 2nd sentence.

8 0

3 years ago

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

amm1812

<u>Answer</u>:

The velocity of the tennis racket after the collision 14.966 m/s.

<u>Step-by-step explanation:</u>

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂² ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

= ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

= 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

7 0

3 years ago

Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg

n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

$\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N$

Thus, the net force over the body is:

$F=(913.14N)\hat{i}+(274.87N)\hat{j}$

Next, you calculate the magnitude of the force:

$F=\sqrt{(913.14N)+(274.87N)^2}=953.61N$

and the direction is:

$\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°$

7 0

3 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago

Someone please quickly help me with this problem?

Crazy boy [7]

The answer is 60 km. I hope it helps i dont know if this is right or wrong.

5 0

3 years ago

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