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2 years ago

You weigh yourself at the top of a high mountain and the scale reads 728 N. If your

Physics

1 answer:

SpyIntel [72]2 years ago

4 0

The gravitational strength is equal to $9.71\frac{N}{kg}$

Why?

Since we are given the gravity force acting (728N) and the mass (75Kg), we can calculate the gravitational strength using the following formula:

$gravityforce=mass*gravitationalStrength\\\\gravitationalStrength=\frac{gravityforce}{mass} \\\\gravitationalStrength=\frac{728N}{75kg}=9.71\frac{N}{kg}$

Have a nice day!

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a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) On doubling the mass:

New mass, $m'=2m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

Then the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

7 0

2 years ago

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co

natima [27]

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L. E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses. (L always stays the same.)

Take the average of your k values.

5 0

2 years ago

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In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

noname [10]

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

6 0

2 years ago

The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is

Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

W = ΔEm

Where the work is

W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

Em₀ = K = ½ m v²

Final. When the system is at rest

$Em_{f}$ = $K_{e1}b +K_{e2}$ = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

∑ F = F - fr

Axis y

N- W = 0

N = W

The friction force has the equation

fr = μ N

fr = μ W

The job

W = (F – μ W) x

We substitute in the equation

(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values and solve

½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

x² 30 + 14.4 x - 1,125 = 0

x² + 0.48 x - 0.0375 = 0

We solve the second degree equation

x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

x = [-0.48 ± 0.617] / 2

x₁ = 0.0685 m

x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

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2 years ago

13) Can two nonzero perpendicular vectors be added together so their sum is zero? Explain.

Neporo4naja [7]

In order for two vectors to add to zero, they must have the same magnitude and point in opposite directions.

Two perpendicular vectors, by definition, make a right angle with each other whereas two vectors pointing in opposite directions form a straight line.

Because of this, two perpendicular vectors with nonzero magnitudes will never add to zero.

3 0

2 years ago

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