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3 years ago

A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the

Physics

1 answer:

Nitella [24]3 years ago

4 0

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

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The density of a substance is defined as its

Fantom [35]

<h2>Answer</h2>

<u>Density is the mass per unit volume</u>

<h2>Explanation</h2>

Density is actually is the ratio of the mass of an object to its volume. It is used to compare the different objects on the basis of their weight. You can say that more the weight of an object more will be its density. For example, the density has the role in floating the ships on the water. When we insert the wooden piece inside water it pops up to the surface. The weight of the wood piece is more than the water and volume is greater which affects its density. When the object gets smaller the size on pressing, the density of the object increase.

7 0

4 years ago

Which value is NOT equivalent to the others?

pishuonlain [190]

Answer:

A) 5 × 1011nm

I think It's A

8 0

3 years ago

"A student bikes to school by traveling first dN = 0.900" and "Similarly, let d⃗ W be the displacement vector corresponding to t

raketka [301]

Complete Question

A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.

Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.

Express your answer as two numbers separated by a comma. Be careful with your signs.

Answer:

The value is $dT = ( -0.3, 0.8)$

Explanation:

From the question we are told that

The first displacement is $dN = 0.900 \ due \ North$ i.e positive y-axis

The second displacement is $dW = 0.300 \ miles \ due \ west$ i.e negative x-axis

The final displacement is $dS = 0.100 \ miles \ due \ south$ i.e negative y-axis

Generally dW in component for is

$dW = (-0.3 , 0)$

Generally the total displacement of the student is mathematically represented as

$dT = ( -0.3, (0.90 - 0.10))$

$dT = ( -0.3, 0.8)$

6 0

3 years ago

An object is hung on the end of a vertical spring and is released from rest with the spring 3 unstressed. If the object falls 3.

den301095 [7]

Answer:

The period of the resulting oscillatory motion is 0.20 s.

Explanation:

Given that,

Spring constant $k= 3\ N/m^2$

We need to calculate the time period

The object is at rest and has no elastic potential but it does has gravitational potential.

If the object falls then the the gravitational potential change in to the elastic potential.

So,

$mgh=\dfrac{1}{2}kh^2$

$m=\dfrac{1}{2}\times\dfrac{kh}{g}$

Where,h = distance

k = spring constant

Put the value into the formula

$m=\dfrac{1\times3\times3.42\times10^{-2}}{2\times9.8}$

$m=5.235\times10^{-3}\ kg$

Using formula of time period

$T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{5.235\times10^{-2}}{3}}$

$T=0.20\ sec$

Hence, The period of the resulting oscillatory motion is 0.20 s.

8 0

3 years ago

The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning

sammy [17]

Let the rise in temperature be $5^0C$

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = $1.7*10^{-5}$ $^0C^{-1}$ , and Δt = $5^0C$

So expansion in length = 93* $1.7*10^{-5}$ *5 = 0.007905 m = $0.79*10^{-3}m$

So order of magnitude in change in length = -3

3 0

3 years ago

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