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Nataly [62]
2 years ago
15

Glacial episodes are an example of

Physics
1 answer:
ira [324]2 years ago
5 0
They are an example of a positive feedback.
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Differentiate the following functions with respect to x <br>xsin x​
gtnhenbr [62]

Answer:

Here is your answer

Hope it helps

6 0
3 years ago
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4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.
VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

v=\dfrac{d}{t}

Where. d = distance

t = time

Put the value into the formula

v=\dfrac{18.288}{2}

v=10\ m/sduring that run

(b). We need to calculate their velocity

Using formula of speed

v=\dfrac{\Delta d}{\Delta t}

Put the value into the formula

v=\dfrac{22.86-36.58}{2}

v=-6.86\ m/s

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

d=vt

Put the value into the formula

d=6.86\times1.3

d=8.91\ m

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
What type of lens does our eye function as, and what is it's purpose?
yKpoI14uk [10]
It is D because our eye lenses reflect the white light we see and it also reflects the light to a point to where we can see colors and objects clearly... Hope this helps out ^-^''
3 0
2 years ago
Read 2 more answers
A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t
Sergio [31]

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

7 0
2 years ago
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