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3 years ago

What is the term used when a ball is hit and the batter reaches the following bases safely (without being called out)?

Physics

2 answers:

trasher [3.6K]3 years ago

8 0

First base would be the term

andre [41]3 years ago

4 0

First base is the answer

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A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the

Svetllana [295]

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$ , b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$ , c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

5 0

3 years ago

Read 2 more answers

Anthony is standing on the top of a building 10 m high holding a 7 kg bowling ball. Mildred dug a 2-m-deep hole next to the base

Leokris [45]

Answer:

Explanation:

8 0

3 years ago

N76 [4]

Answer:

Weight

Explanation:

Weight is the downward pull on an object due to gravity.

For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.

4 0

3 years ago

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a

harina [27]

Answer:

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

Em_f = U = - G Mm / R₂

energy is conserved

Em₀ = Em_f

½ m v_c² - G Mm / R = - G Mm / R₂

v_c² = 2 G M (1 /R - 1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$ )

we make the product and keep the terms linear

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

5 0

3 years ago

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h

QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0

3 years ago