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3 years ago

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A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc

elerates it to the right. It is observed that it takes the box 5.4 seconds to travel 32 meters. What is the magnitude of the force?

1 answer:

iren2701 [21]3 years ago

3 0

Answer:

The correct answer is;

The magnitude of the force is 35.12 N

Explanation:

To solve the question, we note that the friction is zero and the force causes motion of a stationary mass

One of the equations of motion is required such as

v² = u² + 2× a× s

Where

v = Final velocity = 5.93 m/s

u = Initial velocity = 0 m/s , object at rest

a = acceleration

s = distance moved = 32 meters

But v = Distance/Time = 32 m /5.4 s = 5.93 m/s

Therefore

5.93² = 2×a×32

or a = 35.12/ 64 = 0.55 m/s²

Therefore Force F = Mass m × Acceleration a

Where mass m = 64 kg

Therefore F = 64 kg×0.55 m/s² = 35.12 N

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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

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

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

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2 years ago

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

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3 years ago

The elememt sodium (Na) is a ?

luda_lava [24]

It is a very reactive metal with 11 protons ,12 neutrons, 11 electrons, and 1 valence electron

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2 years ago

Read 2 more answers

1.)how is climate different from weather

Zolol [24]

(sorry I can only answer no. 1)
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3 years ago

The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50

sammy [17]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, $\theta_c=52^o$

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

$n_1sin\theta_c=n_2sin\theta_2$

Here, $\theta_2=90$

$sin\theta_c=\dfrac{n_2}{n_1}$

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

$n_1=\dfrac{n_2}{sin\theta_c}$

$n_1=\dfrac{1}{sin(52)}$

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

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3 years ago