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3 years ago

what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2

1 answer:

8 0

The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.

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A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.

kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

Therefore, average speed of the trip = $\frac{\text{Total distance covered}}{\text{Total time taken}}$

From the graph attached,

Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

= 90 meters

Total time taken = 60 seconds

Average speed = $\frac{90}{60}$

= 1.5 meter per second

2). Velocity is a vector quantity.

Therefore, average velocity = $\frac{\triangle d}{\triangle t}$

= $\frac{d_{60}-d_0}{60-0}$

= $\frac{90-0}{60-0}$

= 1.5 meter per second

7 0

3 years ago

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

irinina [24]

The magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

The given parameters;

length of the solenoid, L = 91 cm = 0.91 m
radius of the solenoid, r = 1.5 cm = 0.015 m
number of turns of the solenoid, N = 1300
current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

$B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\$

where;

$\mu_o$ is the permeability of frees space = 4π x 10⁻⁷ T.m/A

$B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T$

Thus, the magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

Learn more here:brainly.com/question/17137684

7 0

2 years ago

A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa

Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute = $W_B$ =10 g

Given mass of solvent= $W_A$ =90 g

$k_f=1.86^{\circ}C/m$

Freezing point of solution =-3.33 $^{\circ}$ C

Freezing point of solvent = $0^{\circ}$ C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33= $3.33^{\circ}$

$\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}$

$M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}$

$M_B=62.06 g/mol$

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0

3 years ago

Read 2 more answers

The magnetic field on the Sun is created by______.

Ber [7]

The answer is actually c hope this helps
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8 0

3 years ago

Question 8

viktelen [127]

Answer: D(t) = $8.e^{-0.4t}.cos(\frac{\pi }{6}.t )$

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = $a.sin(\omega.t)$ or y = $a.cos(\omega.t)$

where:

|a| is initil displacement

$\frac{2.\pi}{\omega}$ is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

$y=a.e^{-ct}.cos(\omega.t)$ or $y=a.e^{-ct}.sin(\omega.t)$

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

$y=a.e^{-ct}.cos(\omega.t)$

period = $\frac{2.\pi}{\omega}$

12 = $\frac{2.\pi}{\omega}$

ω = $\frac{\pi}{6}$

Replacing values:

$D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$

The equation of displacement, D(t), of a spring with damping factor is $D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$ .

3 0

3 years ago