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3 years ago

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Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while

Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

1 answer:

Vinvika [58]3 years ago

7 0

Answer:

Explanation:

Given

Time taken to reach ground is $t=2.7\ s$

Malda initial velocity $u=95\ cm/s$

Let h be the height of Cliff

using $h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

$h=0+\frac{1}{2}(9.8)(2.7)^2$

$h=35.72\ m$

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

$R=u\times t$

$R=0.95\times 2.7=2.43\ m$

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Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1

aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴ (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

3 0

2 years ago

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

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2 years ago

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laila [671]

I’m thinking the Last one tbh

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3 years ago

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a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

Kryger [21]

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒ $v_{rolling}=v_{translational}+v_{rotational}$

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒ $\omega=\frac{v_{translational}}{r}$

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒ $v_{rotational}=\omega \times r$

On substituting the estimated values, we get

⇒ $=\frac{v_{translational}}{r} \times r$

⇒ $=v_{translational}$

So that the velocity will be:

⇒ $v_{rolling}=v+v$

⇒ $=2v$

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3 years ago

A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it mai

Sloan [31]

Answer: Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

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at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa , T = 500°C

h1 = 3434.7 KJ/Kg

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M2 = V2 / v'2

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2 years ago