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3 years ago

a model rocket is launched at an angle of 55 to the horizontal with a momentum of 20 kg-m/. If the vertical component of its vel

ocity is 40 m/s what is the mass of the rocket?

Physics

1 answer:

den301095 [7]3 years ago

4 0

$Given:\\ \alpha =55^\circ\\p=20kg\cdot \frac{m}{s} \\v_v=40 \frac{m}{s} \\\\Find:\\m=?\\\\Solution:\\\\p_v=mv_v\Rightarrow m= \frac{p_v}{v_v} \\\\ \frac{p_v}{p} =\cos \alpha \Rightarrow p_v=p\cos \alpha \\\\m= \frac{p\cos \alpha }{v_v} \\\\m= \frac{20kg\cdot \frac{m}{s} \cdot \cos 55^\circ }{40 \frac{m}{s} } \approx 0.287kg=287g$

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A car decelerates uniformly from 54 m/s to 38 m/s in 8.0 seconds. How far does it travel dunng this time?

galina1969 [7]

Answer:

368m

Explanation:

dx=vt

54+38/2=46

46*8=368m

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3 years ago

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t

soldi70 [24.7K]

Answer:

a) three times greater

Explanation:

Remember the heat capacity is the necessary heat to increase the temperature 1 ºC to an specific mass of the substance, in this case since the water and the metal have the same mass, is easier to compare just the neccesary heat.

Since water increases 20 ºC and the metal 60 ºC, it was neccesary 3 times of the quantity of heat to the same amount of mass of the two different susbtances.

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3 years ago

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A series RLC circuit with R = 15.0 ohms, C = 4.70 uF, and L = 25.0mH is connected to an AC voltage source with V(t) = 75.0 Vrms

Brilliant_brown [7]

(a) The rms current in the circuit is 2.58 A.

(b) The rms voltage of Vab is 38.7 V, Vbc is 158.83 V, Vcd is 222.93 V, Vbd is 64.11 V, and Vad is 75 V.

<h3>Capacitive reactance of the circuit</h3>

The capacitive reactance of the circuit is calculated as follows;

$X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} \\\\X_c = 61.56 \ ohms$

<h3>Inductive reactance of the circuit</h3>

Xl = ωL

Xl = 2πfL

Xl = 2π x 550 x 25 x 10⁻³

Xl = 86.41 ohms

<h3>Impedance of the circuit</h3>

$Z = \sqrt{R^2 + (X_L - X_C)^2} \\\\Z = \sqrt{15^2 + (86.41 -61.56)^2 } \\\\Z = 29.03 \ ohms$

<h3>rms current in the circuits</h3>

$I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{75}{29.03} \\\\I_{rms} = 2.58 \ A$

<h3>rms voltage in resistor (Vab)</h3>

$V_{ab} = I_{rms} R\\\\V_{ab} = 2.58 \times 15\\\\V_{ab} = 38.7 \ V$

<h3>rms voltage in capacitor (Vbc)</h3>

$V_{bc} = I_{rms} X_c\\\\V_{bc} = 2.58 \times 61.56\\\\V_{bc} = 158.83 \ V$

<h3>rms voltage in inductor (Vcd)</h3>

$V_{cd} = I_{rms} X_l\\\\V_{cd} = 2.58 \times 86.41\\\\V_{cd} = 222.93\ V$

<h3>rms voltage in capacitor and inductor (Vbd)</h3>

$V_{bd} = I_{rms} \times (X_l - X_c)\\\\V_{bd} = 2.58 \times (86.41 - 61.56)\\\\V_{bd} = 64.11 \ V$

<h3>rms voltage in resistor, capacitor and inductor (Vad)</h3>

$V_{ad} = I_{rms} \times \sqrt{R^2 + (X_l- X_c)^2} \\\\V_{ad} = 2.58 \times Z \\\\V_{ad} = 2.58 \times 29.03\\\\V_{ad} = 75 \ V$

<h3>Average rate of energy dissipation in the 3 circuit element</h3>

$P = I_{rms}^2 Z\\\\P = (2.58)^2 \times 29.03\\\\P = 193.23 \ W$

Learn more about RLC series circuit here: brainly.com/question/15595203

6 0

2 years ago

A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.

charle [14.2K]

Answer:

(a)F= 3.83 * 10^3 N

(b)Altitude=8.20 * 10^5 m

Explanation:

On the launchpad weight = gravitational force between earth and satellite.

W = GMm/R²

where R is the earth radius.

Re-arranging:

WR² / GM = m

m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg

The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:

Fc = mω²r

where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.

ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s

When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:

Fc = GMm/r²

mω²r = GMm / r²

ω²r = GM / r²

r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²

r³ = 3.612 * 10^20

r = 7.12 * 10^6 m

(a) F = GMm/r²

F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²

F= 3.83 * 10^3 N

(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m

4 0

3 years ago

The dew point is the temperature at which

SIZIF [17.4K]

Dew point is the temperature at which airborne water vapor will condense to form liquid dew.

the most appropriate answer is D !

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3 years ago

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