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3 years ago

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A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 31.0 m/s. A net is positioned a horizont

al distance of 48.5 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

2 answers:

hodyreva [135]3 years ago

5 0

Answer:

s = vcos(x)t

50 = 25cos(45)t

cos(45)t = 2

t = 2/cos(45) = 2sqrt(2)

h = vsin(x)t + gt^2/2

h = 25sin(45)*2sqrt(2) - 4.9*8

h = 10.8 metres

Explanation:

nikklg [1K]3 years ago

4 0

Answer:

24.5 m

Explanation:

First, find the time it takes for the daredevil to travel 48.5 m horizontally:

x = x₀ + v₀ t + ½ at²

(48.5 m) = (0 m) + (31.0 m/s cos 45.0°) t + ½ (0 m/s²) t²

t = 2.21 s

Next, find the vertical position reached at this time:

y = y₀ + v₀ t + ½ at²

y = (0 m) + (31.0 m/s sin 45.0°) (2.21 s) + ½ (-9.8 m/s²) (2.21 s)²

y = 24.5 m

The net should be placed 24.5 meters above the cannon.

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A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

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Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

$v(t)=-gt-v_e\times \ln \frac{m-rt}{m}$

v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

$v(60) = 668.97 m/s$

Height of the rocket = h

$Velocity=\frac{Displacement}{time}$

$668.97 m/s=\frac{h}{60 s}$

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Height of the rocket be one minute after liftoff is 40.1382 km.

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3 years ago

A drunken sailor stumbles 550 meters north, 500 meters northeast, then 450 meters northwest. What is the total displacement and

cluponka [151]

Answer:

Resultant displacement = 1222.3 m

Angle is 88.3 degree from +X axis.

Explanation:

A = 550 m north

B = 500 m north east

C = 450 m north west

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A = 550 j

B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j

C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j

Net displacement is given by

R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j

R = 35.4 i + 1221.8 j

The magnitude is

$R = \sqrt{35.4^{2}+1221.8^{2}}R = 1222.3 m$

The direction is given by

$tan\theta =\frac{1221.8}{35.4}\\\\\theta = 88.3^{o}$

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3 years ago

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scoray [572]

Answer:

read the explanation

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Purchased electricity is fed into our TVs and is converted to light and sound.

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Chemical Energy is converted to Electrical Energy (stove)

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3 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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