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3 years ago

15

Can there be two places on the surface of the Earth having the same local time but opposite seasons? A. No, there cannot be two

such places. B. Yes, only if both of them are on the equator. C. Yes, only if both are in the different hemispheres. D. Yes, only if both of them are at the opposite poles.

1 answer:

Black_prince [1.1K]3 years ago

3 0

Answer : Yes, only if both are in the different hemispheres.

Explanation : At the same local time, the southern hemisphere points away from the sun then creating the winter during June, July and August and when the south pole is tilted toward the sun and the northern hemisphere is tilted away then creating summer in the southern hemisphere is in December , January and February.

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

PtichkaEL [24]

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

m'λ = m"λ'

λ' = m'λ/m"

= 10 × 640 nm/12

= 6400 nm/12

= 533.33 nm

8 0

2 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

During a tornado in 2008 the Peachtree Plaza Westin Hotel in downtown Atlanta suffered damage. Suppose a piece of glass dropped

Darina [25.2K]

Answer:

Time = t = 6.62 s

Explanation:

Given data:

Height = h = 215 m

Initial velocity = $v_{i}$ = 0 m/s

gravitational acceleration = g = 9.8 m/s²

Time = t = ?

According to second equation of motion

$h = v_{i}t + \frac{1}{2}gt^{2}$

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

$h = \frac{1}{2}gt^{2}$

t² = $\frac{2h}{g}$

t = $\sqrt{\frac{2h}{g} }$

t = $\sqrt{\frac{(2)(215) }{9.8} }$

t = 6.62 s

3 0

3 years ago

7. The S.l. unit for force is<br> A. kg<br> C. m/s<br> B. m/s2<br> D. N

netineya [11]

D.N(Newton) this is the S.I unit for force

5 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

6 0

3 years ago

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