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1 year ago

A ball is traveling 24° above the horizontal at a speed of 12 m/s. What is the vertical component of its speed?

Physics

1 answer:

victus00 [196]1 year ago

3 0

Answer:

4.88 m/s

Explanation:

Vertical component would be 12 * sin 24 = 4.88 m/s

Horizontal is 12 * cos 24

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A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

2 years ago

I need help I’ll give you 15 points for both of them

Simora [160]

Answer:

Point c

Explanation:

Because of gravity

5 0

2 years ago

Describe the orbit in which planets revole around the sun

Novay_Z [31]

Planets orbit the sun in the paths which are known as elliptical orbit. Each planet has its own orbit around the sun and direction in which all the planets orbit around the sun are the same. These orbits were well explained by the astronomer Kepler. The gravity of the Sun keeps the planets in their orbits. They stay in their orbits because there is no other force in the Solar System which can stop them.

7 0

1 year ago

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10

bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

$m_1$ = Mass of sphere = 10000 kg

$m_2$ = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

$\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s$

Angular momentum is given by

$L=I\omega$

Moment of inertia of the satellite is

$I_s=\frac{2}{5}m_1r^2$

Moment of antenna of the satellite is

$I_a=\frac{1}{3}m_2l^2$

The angular momentum of the system is

$L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s$

The angular momentum of the satellite is 605447.7066 kgm²/s

5 0

2 years ago

A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional d

Law Incorporation [45]

Answer:

Third displacement = 2.81 m which is 61.70° north of east.

Explanation:

Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

She sails 2 km east, displacement = 2 i

Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis

Displacement = 3.5 cos 315 i + 3.5 sin 315 = 2.47 i - 2.47 j

Let third displacement be x i + y j

We have final displacement = 5.80 km east = 5.80 i

From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j

= (4.47+x) i + (y - 2.47) j

Comparing both , we have 4.47+x = 5.80

x = 1.33

y-2.47=0

y = 2.47 j

So third displacement = 1.33 i + 2.47 j

Magnitude of third displacement = $\sqrt{1.33^2+2.47^2} =2.81m$

θ = tan⁻¹(2.47/1.33) = 61.70°

So third displacement = 2.81 m which is 61.70° north of east.

7 0

2 years ago