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3 years ago

NEED HELP ASAP! DUE NOW!!

Physics

1 answer:

docker41 [41]3 years ago

4 0

Oh okay the first person who answered is incorrect but

0—> PE=0J

1—>PE=1J

20—>PE=20J

50—>PE=50J

100—>PE=100J

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Help with physics projectile motion

BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0

4 years ago

20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

aliina [53]

$Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot \frac{4m^3}{6.5m^3} \cdot \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C$

7 0

3 years ago

As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.

Colt1911 [192]

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is

$f' = \frac{v + v_o}{v+ v_s} f$

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

7 0

3 years ago

Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground

Readme [11.4K]

PE = (mass) (gravity) (height)

PE = (0.005 kg) (9.8 m/s²) (5 m)

<em>PE = 0.245 Joule</em>

4 0

3 years ago

Find equivalent resistance. <br><br>Answer asap and please, please don't spam.

Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

$\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega$

So, the equivalent resistance s 4.77 ohms.

4 0

3 years ago