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rosijanka [135]
2 years ago
15

John Lester is at the park practicing his pitching. Since he doesn’t have a catcher, John wants to pitch the ball all the way ar

ound the Earth so he can catch it himself. Neglecting air drag and any obstacles that might be in the way, about how fast must the ball be moving? You may take the radius of the Earth to be 6400 km.
Physics
1 answer:
Lisa [10]2 years ago
4 0

Answer:

v_{o} = 79.314\times 10^{6}\,\frac{m}{s}

Explanation:

Let assume that John Lester has a height of 1.80 meters and throws the ball at 70 percent of John Lester's height. The time before the ball hits the soil is:

0\,m = 0.7\cdot (1.80\,m) -\frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}

t \approx 0.507\,s

The initial horizontal velocity required to pitch the ball all the way around the Earth is:

2\pi\cdot (6.4\times 10^{6}\,m)= v_{o}\cdot (0.507\,s)

v_{o} = 79.314\times 10^{6}\,\frac{m}{s}

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A 10 kg ball moving at 13 m/s strikes a 20 kg ball at rest. after the collision the 10 kg ball is moving with a velocity of 7m/s
Lostsunrise [7]

Answer:

30 (kg)

Explanation:

therefore the mass of the ball is 2 so 30 (kg)

6 0
2 years ago
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A conducting rod (length = 80 cm) rotates at a constant angular rate of 15 revolutions per second about a pivot at one end. A un
Studentka2010 [4]

Answer:

3.62 V

Explanation:

L = 80 cm = 0.8 m

f = 15 rps

B = 60 m T = 0.060 T

ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s

v = r ω

here, r be the radius of circular path. Here r = length of rod = L

v = 0.80 x 94.2 = 75.36 m/s

The motional emf is given by

e = B  v  L = 0.060 x 75.36 x 0.8 = 3.62 V

4 0
2 years ago
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts
tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

now in Y direction we can use kinematics

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

4 0
2 years ago
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