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Aliun [14]
2 years ago
6

Which of the following is not part of a circuit? film Load Key Cell

Physics
1 answer:
qaws [65]2 years ago
8 0

Explanation:

I believe <u>f</u><u>i</u><u>l</u><u>m</u><u> </u><u>i</u><u>s</u><u> </u>not part of a circuit

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Which statement represents how light travels? Light given off from a moving source will travel faster in a vacuum. Light given o
Debora [2.8K]

Answer:

the first one

Explanation:

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2 years ago
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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m
Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

V_{escape}=\sqrt{\frac{2GM}{R}}

V_{escape}= Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}

           =15.88 \times 10^4 m/s

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

           

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It's always a good idea to wear a seatbelt in a car, because if the car comes to a sudden stop, you will not move forward very much since the seatbelt is holding you back. The answer is letter D. 

4 0
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