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Tomtit [17]
3 years ago
5

Select all that apply. PLZ

Physics
1 answer:
Crank3 years ago
5 0
A and c

do not complete the circuit there is no battery on a and paper dose not conduct electricity
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A nail gets magnetized when kept close to a magnet
avanturin [10]
In unmagnetized materials the domains are randomly aligned in different directions and cancel each other out. In a magnet all the domains are oriented in the same direction. In the case of a nail, the domains can be aligned in the same direction causing the nail to become magnetic.
3 0
4 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
A bodybuilder deadlifts a 215 kg weight to a height of 0.90 m above the ground. If he deadlifts this weight 10 times in a span o
wariber [46]

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

<h3>What is power?</h3>

In physics, power (P) is the work (W) done over a period of time.

  • Step 1. Calculate the work done by the bodybuilder each time.

The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.

W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N

  • Step 2. Calculate the work done by the bodybuilder over 10 times.

W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N

  • Step 3. Calculate the power exerted by the bodybuilder.

The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.

P = 1.9 × 10⁴ N/45 s = 421 W

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

Learn more about power here: brainly.com/question/911620

#SPJ1

4 0
2 years ago
Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

where

\Delta U is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

Q=500 J

therefore, the work done by the system is

W=Q-\Delta U=500 J-47 J=453 J

5 0
3 years ago
Read 2 more answers
An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met
ASHA 777 [7]

Answer:

B) 2g

Explanation:

<u>Given the following data;</u>

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Acceleration, a = \frac {14^{2}}{10}

Acceleration, a = \frac {196}{10}

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

4 0
3 years ago
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