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3 years ago

Look at this picture of a frog.

Physics

2 answers:

ivanzaharov [21]3 years ago

7 0

It’s 100% biosphere

fredd [130]3 years ago

6 0

Answer:

Biosphere

Explanation:

The biosphere consist of all living organisms

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A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t

Crank

Answer:

$Hmax=15.06$ meters

Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function

The maxima of the function occurs when the slope is zero. i.e.

$\frac{df}{dt} =0\\\frac{df}{dt} =\frac{d}{dt} (-4.9t^2+16t+2)\\\frac{df}{dt} =-4.9*2t+16\\-9.8t+16=0\\t=16/9.8\\t=1.63 secs$

Hence the maxima occurs at t=1.63 seconds

The maximum value of f is

$f(1.63)=-4.9(1.63^2)+16(1.63)+2\\f(1.63)=15.06\\$

hence maximum height is found to be

$Hmax=15.06$ meters

8 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 2.52 m/s. (a) After 4.00 s, what is the speed of th

gogolik [260]

Answer: 41.72m/s

Explanation: v= u + gt

V = 2.52 + 9.8(4.00)

V = 41.72m/s

6 0

2 years ago

A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00

Contact [7]

Answer:

2 m/s

Explanation:

$m_1$ = Mass of red truck = 1000 kg

$m_2$ = Mass of green truck= 3000 kg

$u_1$ = Initial Velocity of red truck = 6 m/s

$u_2$ = Initial Velocity of green truck

$v$ = Velocity with which they move together = 0

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s$

Velocity of the green truck is 2 m/s

8 0

2 years ago

- Contemporary World Issues -

kupik [55]

Answer:

hey wsg

Explanation:

8 0

2 years ago