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Ivan
3 years ago
11

A pitched ball is hit by a batter at a 45degrees angle and just clears the outfield fence, 98m away. Assume that the fence is at

the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
Physics
1 answer:
Alex17521 [72]3 years ago
8 0
The range of a projectile can be found directly using:
R = (v²sin2∅) / g
v = √((98 x 9.81)/(sin(90)))
v = 31.0 m/s
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In levelling, the following staff readings were observed involving an inverted staff, A = 2.915 and B = -2.028. What is the rise
salantis [7]

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

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3 years ago
If a wave has a frequency of 50 Hz and a wavelength of 3 meters, what is its Velocity? (Use
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Answer:

the answer is 150m/s

Explanation:

v=λf

v=50*3

v=150m/s

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3 years ago
Why do remote controls for TV’s use infrared waves to communicate?
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Answer:  An IR remote (also called a transmitter) uses light to carry signals from the remote to the device so it can be controlled. It emits pulses of invisible infrared light that correspond to specific binary codes. These codes represent commands, such as power on, volume up, or channel down.

Explanation:

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2 years ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
3 years ago
Rank the following radiation particles according to their ability to penetrate solid objects, from the type of particle best abl
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Rank of the radiation particles according to their ability to penetrate solid objects, from the best able to penetrate to the least able to penetrate:
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3 years ago
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