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3 years ago

Back to 'Energy & Work'

Physics

1 answer:

mixer [17]3 years ago

3 0

Answer:

Explanation: Two possible solutions

in the absence of friction

mgh = ½mv²

h = v²/2g = 24.4²(2(9.81) = 30.344... = 30.3 m

if enough rolling friction exists to keep the ball from sliding, but ignoring air resistance and assuming the ball is already rolling smoothly at the start

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½((2/3)mR²)(v/R)²

2gh = v² + (2/3)v²

2gh = (5/3)v²

h = (5/6)v²/g

h = (5/6)24.4²/9.81 = 50.5742 = 50.6 m

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A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

sergij07 [2.7K]

To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.

First, we manipulate the one of the kinematic equations

v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s

7 0

3 years ago

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

$m_1v_1+m_2v_2=(m_1+m_2)V$

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

$v_2=-\dfrac{m_1v_1}{m_2}$

$v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}$

$v_2=-1.47\ m/s$

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0

3 years ago

light waves with a frequency of 7.26 x 10^14 HZ lies in the violet region of the visible spectrum. what is the wavelength of thi

pogonyaev

Answer:

4.13×10⁻⁷ m

Explanation:

Applying

v = λf................... Equation 1

Wherfe v = speed of light, λ = wavelength of light, f = frequency of light.

make λ the subject of the equation

λ = v/f....................... Equation 2

Note: Light is an electromagnetic wave, and all electromagnetic wave moves with thesame speed which is 3×10⁸ m/s.

From the question,

Given: f = 7.26×10¹⁴ Hz

Constant: v = 3×10⁸ m/s

Substitute these values into equation 2

λ = (3×10⁸)/(7.26×10¹⁴)

λ = 4.13×10⁻⁷ m

7 0

3 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

7 0

4 years ago

0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre

Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e² = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0

3 years ago