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2 years ago

HELP PLEASE!Can anyone help with any of these?

Physics

1 answer:

nadezda [96]2 years ago

6 0

1. The magnitude is 49.00 meters and his direction is 7.43 degrees to the right of his original position.

2. D = 7.7 km at 26.2° to ground ANS

3. 172 km

4. 172 km in a direction 34.2 degrees East of North

<em>I hope this helps you!!!</em>

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What is the best way to avoid being run over by your own pwc or motorboat?

kobusy [5.1K]

Wear an ignition safety switch lanyard so you can turn it off. Most powerboats and PWCs come furnished by the maker with a crisis motor cut-off switch. This wellbeing gadget can stop the motor if the administrator tumbles off the PWC or out of the powerboat, or is generally tossed from the best possible working position.

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2 years ago

What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.

julia-pushkina [17]

<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Finding the velocity from (1):

$V=\frac{p}{m}$ (2)

$V=\frac{4000kg.m/s}{115kg}$

<u>Finally:</u>

$V=34.78m/s$ >>>This is the velocity of the object

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2 years ago

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In the equation for the gravitational force between two objects, which quantity must be squared?

Trava [24]

The distance quantity/ measurement must be squared.

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2 years ago

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A 90kg mountain climber hangs from a nylon rope and streches it ny 25.0cm. if the rope was originally 30.0m long and it's diamet

bixtya [17]

Answer:

Explanation:

E = σ/ε = (F/A) / (ΔL/L)

E = (mg/(πd²/4) / (ΔL/L)

E = (4mg/(πd²) / (ΔL/L)

E = 4Lmg/(πd²ΔL)

E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)

E = 1.35 x 10⁹ Pa or 1.35 GPa

7 0

1 year 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.

6 0

2 years ago