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xeze [42]
3 years ago
13

A golf ball is hit with an initial velocity of 155 ft/s at an angle of 23 degrees with the horizontal. After 1.4 seconds, how fa

r has the golf ball traveled horizontally and vertically?. a. 199.7 feet horizontally and 75.2 feet vertically. b. 60.6 feet horizontally and 168.4 feet vertically. c. 199.7 feet horizontally and 53.4 feet vertically. d. 60.6 feet horizontally and 174.4 feet vertically.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

<span>2) s(t) = 155tcos(23°) </span>
<span>=> s(1.4) = 155 * 1.4 * cos(23°) </span>
<span>=> s(1.4) ≈ 199.7 ft (1 dp) <=- distance displaced horizontally </span>

<span>a(t) = -32 ft/s^2 </span>
<span>=> v(t) = -32t + 155sin(23°) </span>
<span>=> h(t) = -16t^2 + 155tsin(23°) </span>
<span>=> h(1.4) = -16(1.4)^2 + 155(1.4)sin(23°) </span>
<span>=> h(1.4) ≈ 53.4 ft (1 dp) <== distance displaced vertically.</span>
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Deffense [45]

Answer:

The angle of launch is 52.49 Degree.

Explanation:

The Range R and Height H of a thrown object is calculated using the formula,

R=V₀² sin(2φ)/g

H=V₀²sin²(φ)/g

From these equations it can be written,

V₀²=R g/ sin(2φ)

V₀²=H g/ sin²(φ)

These values are equal so it can be written by equating these equations,

R g/sin(2φ)=H g/sin²(φ)

tan(φ)= 2H/R

Given H=72.3 m and R=111 m, the angle of launch is,

tan(φ)= 2*72.3/111

φ= 52.49 Degree.

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2 years ago
An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
Wewaii [24]

Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

P = 170 W

now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

u = \frac{B^2}{2\mu_0}

intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

4 0
3 years ago
Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part
sergey [27]

Answer:

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and  L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

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3 years ago
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ra1l [238]

Answer:

1) The plane of the loop is perpendicular to the magnetic field.

2) The magnetic flux is independent of the orientation of the loop.p

Explanation:

The flux is calculated as φ=BAcosθ. The flux is therefore the highest when the magnetic field vector is perpendicular to the plane of the loop We can also deduce that the flux is zero when there is no magnetic field part perpendicular to the loop When the angle reaches zero, the flux is in the limit because when the angle becomes zero, the cos is the maximum.

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Rus_ich [418]

Answer:

2.3  ×  10^{-1}  

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

4 0
3 years ago
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