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3 years ago

A velocity-time graph is shown below: (2 points)

Physics

1 answer:

BabaBlast [244]3 years ago

5 0

Answer:

2m/s²

Explanation:

V=Vstart+at
rewrite that to find a so a=(V-Vstart)/t

part A of graph

a=(20m/s–0m/s)/5s
a=4m/s²

part B of graph

a=(0m/s–0m/s)/5s
a=0m/s²

the average between the two is both answers added divided by the number of answers

(4m/s²–0m/s²)/2
4m/s²/2
2m/s²

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What is the natural frequency (wo) of a circuit that has an inductor of 0.048 henery's, and a capacitor of 0.0004 farads?

adell [148]

Answer:

The natural frequency will be 228.11 rad/sec

Explanation:

We have given Inductance L = 0.048 Henry

And the capacitance C = 0.0004 farad

We have to find natural frequency

When only inductor and capacitor is present in the circuit then it is known as LC circuit and Natural frequency of LC circuit is given by $\omega _0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.048\times 0.0004}}=228.21\ rad/sec$

So the natural frequency will be 228.21

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3 years ago

Define the focus of a concave lens

Sophie [7]

Answer:

Principal focus of concave lens - 

★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.

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• Additional information - 

★ Principal focus - A number of rays parallel to the principal axis after reflection from a concave mirror meet at a point on the principal axis or appear to come from a point after reflection from a convex mirror on the principal axis. This is called principal focus.

8 0

3 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

Which location, 23 degrees or 48 degrees would experience the same earthquake at stronger intensity?Explain why.

BaLLatris [955]

Answer:

48 degress

Explanation:

An earthquake causes many different intensities of shaking in the area of the epicenter where it occurs. So the intensity of an earthquake will vary depending on where you are. Sometimes earthquakes are referred to by the maximum intensity they produce. In the United States, we use the Modified Mercalli Scale. Earthquake intensity is a ranking based on the observed effects of an earthquake in each particular place. Therefore, each earthquake produces a range of intensity values, ranging from highest in the epicenter area to zero at a distance from the epicenter.

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3 years ago

An astronaut is in space with a baseball and a bowling ball. The astronaut pushes both objects in the same direction. If both ba

Crazy boy [7]

Answer:

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

Explanation:

This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is

p = m v

Besibol Blade

p1 = m1 v

Bowling ball

p2 = m2 v

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

p2 >> p1

3 0

3 years ago