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3 years ago

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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav

es the bat traveling to the left at an angle of 30 ∘ above horizontal with a speed of 65.0 m/s . The ball and bat are in contact for 1.75 ms .

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

Final speed of the baseball, $v=65\ cos(30)=56.29\ m/s$

Time of contact, $t=1.75\ ms=1.75\times 10^{-3}\ s$

(a) It is assumed to find the horizontal component of average force. It is given by :

$F=m\dfrac{v-u}{t}$

$F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}$

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

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Visible light waves do not diffract as well as radio waves because

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they have more energy than radio waves.

&

because the wavelength of the light waves are too small

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3 years ago

Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ram

Sonbull [250]

<h2><u>Full Question:</u></h2>

Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?

answer choices

Marble 1 has a faster speed than Marble 2.

Marble 2 has a faster speed than Marble 1.

Both the marbles travel at the same speed.

There is not enough data to compare the speeds of marbles.

<h2><u>Answer:</u></h2>

Marble 2 has a faster speed than Marble 1.

Option B.

<h3><u>Explanation:</u></h3>

The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.

Time taken by the ball = 10 seconds.

Distance covered by 1st ball = 20 cm.

Distance covered by 2nd ball = 3cm.

So speed of the 1st ball = 2cm/sec.

Speed of the 2nd ball = 3 cm /sec.

So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.

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Answer:

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Heat released by the largest container is

Q_a = M ce (T_{i}-T_{f})

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m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

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