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3 years ago

10

50 POINTS! Chipping from the rough, a golfer sends the ball at an angle of 54° with an initial velocity of 13m/s, how far down t

he green did the ball go?

1 answer:

docker41 [41]3 years ago

5 0

Range equation:

$x=\dfrac{{v_i}^2\sin2\theta}g$

This ball travels a distance of

$x=\dfrac{\left(13\frac{\rm m}{\rm s}\right)^2\sin108^\circ}{9.8\frac{\rm m}{\mathrm s^2}}=16\,\mathrm m$

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A 100 kg boat is floating in water. half of the boat is submerged under water. what is the weight of the boat?

Alex17521 [72]

W=MG
w is weight
m is mass
g is gravity
W=(100 kg)(9.8 m/s)
W= 980 N
hope this helps

7 0

3 years ago

Why does a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude

inna [77]

Answer:

According to your question although I think an object undergoing uniform circular motion is moving with a constant speed. Nevertheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards,therefore a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude.

3 0

4 years ago

A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

$\mu = \frac{m}{L}$

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

$\mu = \frac{0.09\ kg}{1\ m}$

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

7 0

3 years ago

A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic

defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0

3 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

<h2> $T^{2} =\frac{4\pi^{2}}{GM}a^{3}$ (1) </h2>

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

<h2> $T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2) </h2>

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

<h2> $T=152938.0934s$ (3) </h2>

Which is the same as:

<h2> $T=42.482h$ </h2>

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

4 years ago