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3 years ago

A snail can move approximately.3 meters per minute. How many meters can a snail travel in 45min?

Physics

1 answer:

Aloiza [94]3 years ago

6 0

Answer:

13.5

Explanation:

.3×45=13.5 See that is correct u can check with a calculator, I used a calculator

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Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway

SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s$

So, the velocity of the cart is 12.12 m/s.

7 0

3 years ago

How do you do to your question to see if its answered?

Jobisdone [24]

Answer:

Go in notifications, it'll show if it was answered. If it doesn't show that a person answered it, wait a while, someone might respond :)

Explanation:

On the Top right of your screen, theres a bell button. Click that and it will show all the notifications. it will also show if a person answered it.

It will pop up like

*random username* answered your question! ]

Hope this helped

6 0

3 years ago

Read 2 more answers

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

3 years ago

ivann1987 [24]

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

4 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

3 years ago