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3 years ago

You push a 28kg wheelbarrow forward at a constant velocity with 600 N of force. How much is friction pushing backwards?

Physics

1 answer:

AlladinOne [14]3 years ago

7 0

F = applied force on the wheelbarrow in forward direction = 600 N

f = frictional force pushing in backward direction of wheelbarrow = ?

m = mass of the wheelbarrow = 28 kg

Since wheelbarrow moves forward at a constant velocity

a = acceleration of the wheelbarrow in forward direction = 0 m/s²

force equation for the motion of wheelbarrow in forward direction is given as

F - f = ma

inserting the above values in the formula

600 - f = (28) (0)

600 - f = 0

f = 600 N

hence the frictional force comes out to be 600 N

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Assume that the shape of Virginia beach is rectangular.

Note that
1 km = 10³ m
1 cm = 10⁻² m

The area is
A = (15 km)*(50 km)
= (15 x 10³ m)*(50 x 10³ m)
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Because 2 cm of rain fell, the volume is
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Answer: 1.5 x 10⁶ m³

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3 years ago

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th

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Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

$v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}$

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

$v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}$

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

$v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s$

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

brainly.com/question/7538238

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2 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

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A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

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4 years ago

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Crazy boy [7]

Answer:

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Explanation:

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