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skad [1K]
3 years ago
15

Gumuhit ng paglalaruan ngagawang sulok at ibigay ang bawatsukat nito.​

Physics
1 answer:
Anarel [89]3 years ago
8 0

Answer:

there's a link to answer

Explanation:

there's a link to answer in another's persons response it helps a lot just click on it

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A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After
svetlana [45]

Explanation:

Below is an attachment containing the solution.

6 0
4 years ago
A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.
Alex73 [517]

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

                                 d' = d cos Ф

                                     = 1430 cos 5.76°

                                     = 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

5 0
3 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
3 years ago
A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
GalinKa [24]

Answer:

R = 98304.75 m = 98.3 km

Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

Density = 1 x 10⁹ kg/m³

Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

Therefore,

1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]

4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

R = ∛[0.95 x 10¹⁵ m³]

<u>R = 98304.75 m = 98.3 km</u>

6 0
3 years ago
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