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3 years ago

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s

1 answer:

Burka [1]3 years ago

8 0

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

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Speed is the rate something is moving, while velocity is the rate something is going in a given direction.

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linear charge density of system of two line charges is given as

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now we can find the charge by

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3 years ago

If there is 8 g of a substance before a physical change , how much will there be afterwards?

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d = Distance between plates = 2 cm (assumed)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric flux is given by

$\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2$

The charge per unit area on the plates is $1.77\times 10^{-7}\ C/m^2$

Deflection is given by

$s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m$

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3 years ago

An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above

s344n2d4d5 [400]

Answer:

a) No

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Explanation:

Given:

Height of the wall = 3.95m

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Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

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3 years ago