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3 years ago

A vector is 14.4 m long and

Physics

1 answer:

Vaselesa [24]3 years ago

8 0

The y-component of the vector if the vector is 14.4 m long and points in a 133 degree direction is 9.92m

A vector force is defined as a vector quantity having both magnitude and direction.

If the magnitude of a vector is 14.4m long points in a 133-degree direction, then the y-component of the vector force will be expressed as:

$v_y = vsin \theta$

Given the following parameters

$v = 14.4m\\\theta = 133-90=43^0$

Substitute the given parameters into the expression

$v_y=14.4 sin 43^0\\v_y=9.82m$

Hence the y-component of the vector if the vector is 14.4 m long and points in a 133 degree direction is 9.92m

Learn more here: brainly.com/question/24629099

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A series RL circuit with L = 3.00 H and a series RC circuit with C = 3.00 F have equal time constants. If the two circuits conta

Tems11 [23]

Answer:

(a) R = 1Ω

(b) τ = 3

Explanation:

The time constants of the given circuits are as follows

$\tau_{RL} = \frac{L}{R}\\\tau_{RC} = RC$

If the two circuits have equal time constants, then

$\frac{L}{R} = RC\\R^2 = \frac{L}{C} = \frac{3}{3} = 1\\R = 1\Omega$

Therefore, the time constant in any of the circuits is

$\tau = RC = 3$

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3 years ago

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

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4 years ago

How much work is done if you push a box 150 meters with a force of 1.4 N

ss7ja [257]

The work done to push the box is equal to the product between the force and the distance through which the force is applied:
$W=Fd$
In our problem, the force is F=1.4 N and the distance covered is d=150 m, so the work done by pushing the box is
$W=Fd=(1.4 N)(150 m)=210 J$

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4 years ago

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iren [92.7K]

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therefore;
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16960 = 2890 V
V = 5.869 m/s
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3 years ago

Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz

asambeis [7]

By using the Plancks-Einstein equation, we can find the energy;
E = hf
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E = (6.63 x 10⁻³⁴) x (3.55 x 10¹⁷)
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3 years ago