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2 years ago

Question 10 of 34

1 answer:

3 0

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

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A mouse is running along the floor in a straight line at 1.3 m/s. A cat runs after it and, perfectly judging the distance d to t

sukhopar [10]

Answer:

Option b

Solution:

As per the question:

Speed of the mouse, v = 1.3 m/s

Speed of the cat, v' = 2.5 m/s

Angle, $\theta = 38^{\circ}$

Now,

To calculate the distance between the mouse and the cat:

The distance that the cat moved is given by:

$x = v'cos\theta t$

$x = 2.5cos38^{\circ}\times t = 1.97t$

The position of the cat and the mouse can be given by:

$x = x' + vt$

$1.97t = x' + 1.3t$

x' = 0.67 t (1)

The initial speed of the cat ahead of the mouse:

u = $v'sin\theta = 2.5sin38^{\circ} = 1.539\ m/s$

When the time is 0.5t, the speed of the cat is 0, thus:

$0 = u - 0.5tg$

$t = \frac{1.539}{0.5\times 9.8} = 0.314\ s$

Substituting the value of t in eqn (1):

x' = 0.67(0.314) = 0.210 m

Thus the distance comes out to be 0.210 m

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3 years ago

The diagram shows a motion map

Goshia [24]

Answer:

the object has decreasing acceleration and increasing velocity

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3 years ago

Evaluate the energy of the blackbody photons inside your eye. Compare this with the visible energy inside your eye while looking

evablogger [386]

Answer

given,

the volume of eye

$V = \dfrac{4}{3}\pi r^3$

$V = \dfrac{4}{3}\pi 1.5^3\times 10^{-6}$

$V =4.5 \times 10^{-6}$

temperature 37 = 273 + 37 = 310 K

Radiation energy density

$u = aT^4$

$u = \dfrac{4\sigma }{cT^4}$

energy of the BB photons

$E = \dfrac{\sigma }{cT^4V}$

$E = 9.9 \times 10^{-11}\ J$

Flux from the lamp

$F = \dfrac{100}{4\pi \times \pi \times (0.1 \times 10^{-2})^2}$

$F = 2.5 \times 10^{-5} J/s$

Full energy of lamp

$N = \dfrac{F\times 2 R_{eye}}{c}$

$N = \dfrac{2.5 \times 10^{-5}\times 0.03}{3\times 10^{10}}$

$N = 2.5 \times 10^{-15} J$

when we close your eyes the light from the source stop and the black body photon starts hitting retina which will cause darkness.

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3 years ago

Qual unidade de midia usado pelo si no tempo

SIZIF [17.4K]

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3 years ago

In Fig. 7-30, a block of ice slides down a frictionless ramp at angle 50 while an ice worker pulls on the block (via a rope) wit

pentagon [3]

Answer:

It Is 25J greater.

Explanation:

Applying the Energy conservation theorem, we have:

$K_i+U_i+W_{ext}=K_f+U_f\\W_{ext}=\Delta K+\Delta U$

The Work done by the ice worker is given by:

$W_{ext}=F_w*d*cos(\theta)\\W_{ext}=50N*0.50m*cos(180)\\W_{ext}=-25J$

The angle is 180 degrees because the force exerted by the worker is opposite to the motion.

$\Delta U=\Delta K-W_{ext}\\\Delta U=80J-(-25J)\\\Delta U= 105J$

If the rope was not there, the external work is 0, so:

$\Delta U=\Delta K$

$\Delta k=105J$

The kinetic energy is 25J greater without the rope.

$\Delta K_f=105J-80J=25J$

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3 years ago