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3 years ago

OMG PLZ PLZ HELP!!

Physics

2 answers:

horsena [70]3 years ago

4 0

1,3,4 I’m pretty sur e

Ede4ka [16]3 years ago

4 0

Answer:

1,4,5 is right

Explanation:

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D. If a dog has a mass of 12 kg, what is its weight on Neptune?<br> 11.7N/kg

steposvetlana [31]

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

4 0

3 years ago

Add a small ball to a graduated cylinder containing 10 milliliters of water.

malfutka [58]

The volume of water will increase . If yu subtract the original volume from the new volume of water you will get the volume of the small ball.

8 0

3 years ago

Are there any exceptions to the rule that planets rotate with small axis tilts and in the same direction as they orbit the sun?

Alona [7]

<span>Venus, Uranus, and Pluto are exceptions</span>

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3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago