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kakasveta [241]
2 years ago
13

Question 22

Physics
1 answer:
N76 [4]2 years ago
5 0

Answer:

tan theta = -18 / 25 = -35.8 deg

This direction is south of east

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The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?​
Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}

Put all the values,

h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m

So, it will reach to a height of 2.5 m.

8 0
3 years ago
A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12
Tanya [424]

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

20.28\ \text{m}

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

8 0
3 years ago
The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp
Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

  • Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

  • A is the amplitude = 0.5 m
  • ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s  = 0.786 \ m/s

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

5 0
2 years ago
What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
lora16 [44]

Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or

<em>ρ</em> = <em>m</em> / <em>v</em>

Solving for <em>v</em> gives

<em>v</em> = <em>m</em> / <em>ρ</em>

So the given object has a volume of

<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³

5 0
3 years ago
An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0
Diano4ka-milaya [45]
The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = <span>µ x n x I where:
</span>µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = <span>µ </span><span>o = 4 pi x 10^-7 Tm/A
</span>I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT
5 0
3 years ago
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