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2 years ago

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A ball is thrown with velocity 30.m⋅s−1at an angle of 40.º below the horizontal. What is the magnitude of the horizontal compone

nt of the velocity? Give your answer to the nearest m⋅s−1 and do not include units with your answer.

1 answer:

Sophie [7]2 years ago

4 0

Answer:

Explanation:

The horizontal component of the velocity is expresssed as;

Vx = Vcos θ

V is the velocity of the ball

θ is the angle of inclination to the horizontal

Given

V = 30m/s

θ = 40°

Vx = 30cos40°

Vx = 22.98

Hence the magnitude of the horizontal component of the velocity to the nearest m/s is 23

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A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?

andreev551 [17]

<span>Hello!

We have the following data:
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Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

$P = \frac{E}{T}$

Solving:

$P = \frac{E}{T}$

$P = \frac{E}{T} \to T = \frac{E}{P}$

$T = \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}$

$\boxed{T = 3000\:seconds}$

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

$\frac{1}{y} = \frac{60}{3000}$

<span>Product of extremes equals product of means
</span>
$60*y = 1*3000$

$60y = 3000$

$y = \frac{3000}{60}$

$\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark$

I hope this helps! =)
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2 years ago

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A child blows a leaf straight up in the air from rest. The leaf accelerates at 1.0\,\dfrac{\text m}{\text s^2}1.0 s 2 m1, point,

Neko [114]

Answer:

Time taken by the leaf to displace by 1.0 m distance is

$t = \sqrt2$ seconds

Explanation:

As we know that initial velocity of the leaf is given as

$v_i = 0$

now the acceleration upwards for the leaf is

$a = 1 m/s^2$

The displacement of leaf in upward direction is

d = 1 m

so now we have

$d = v_i t + \frac{1}{2}at^2$

$1 = 0 + \frac{1}{2}(1) t^2$

$t = \sqrt2$ seconds

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2 years ago

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A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$

$v_{i}$ is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

$\Delta x = 5(6) - \frac{4}{2}.6^{2}$

$\Delta x = 30 - 2.36$

$\Delta x$ = - 42

Displacement of the boat for t=6.0s interval is $\Delta x$ = - 42m, i.e., 42 m to the left.

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Sladkaya [172]

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2 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

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2 years ago