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elixir [45]
3 years ago
13

1

Physics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

(3) 180 m

Explanation:

y = y₀ + vt − ½ at²

where y is the final height,

y₀ is the initial height,

v is the final velocity,

a is acceleration,

and t is time.

Given y₀ = 0 and a = -10.

y = vt + 5t²

At maximum height, v = 0:

y = 5t²

The ball reaches the same height at 4s and 8s, so the peak is in the middle at 6s.

y = 5(6)²

y = 180

The ball reaches a maximum height of 180 m.

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Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with
IgorLugansk [536]

(a) 2.5\cdot 10^{-6}eV

The energy of a photon is given by:

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the microwave photon,

\lambda=50.00 cm = 0.50 m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV

(b) 2.5 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the visible light photon,

\lambda=500 nm = 5 \cdot 10^{-7}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV

(c) 2500 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the x-ray photon,

\lambda=0.5 nm = 5 \cdot 10^{-10}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV

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Don't worry eventually it is going to stop somewhere or land right
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Bar magnets placed on table may not point along N-S direction . Why? Give reasons.
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Answer:

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3 years ago
Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c
jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

           $=-17850 - 30018.6$

           $=-47868.6 \ Pa$

           $=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

         

4 0
3 years ago
A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m
mestny [16]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

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