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3 years ago

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A football player is running with a velocity or 10 m/s. At that velocity, his momentum is 2500 kg*m/s. What is the football play

ers mass?

1 answer:

DedPeter [7]3 years ago

7 0

The mass of the football player is 250 kg.

<u>Explanation:</u>

Momentum is defined as the product of mass and velocity. So here the velocity (v) is given as 10 m/s and the momentum is given as 2500 kg m /s. So we can determine the mass (m) of the player by substituting the known terms in the formula of determining momentum as shown below.

$\text {Momentum}(p)=m \times v$

As we know the value of momentum and velocity, the mass can be found as,

$m=\frac{p}{v}=\frac{2500}{10}=250 \mathrm{kg}$

Thus, the mass of the football player is found to be 250 kg.

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In order to slide a heavy desk across the floor at constant speed in a straight line, you have to exert a horizontal force of 40

san4es73 [151]

Answer:

F = f from Newton’s first law.

Explanation:

since the desk is moved in a straight line at a constant speed, newton first law tell us that the two forces must be equal.

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. since the table has been set in motion by the 400 N force, it will remain in motion unless it is been acted upon by an external force, and this means that the 400 N must be equal to the frictional force for it to have been in motion in the first instance.

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3 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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3 years ago

Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

cupoosta [38]

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

$\mathbf{v_f ^2 = v_i^2 + 2gh}$

$\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}$

$\mathbf{v_f ^2 =151 m/s}$

$\mathbf{v_f =\sqrt{151 m/s}}$

$\mathbf{v_f =12.29 \ m/s}$

From the equation of continuity:

$\mathbf{A_iV_i = A_fV_f}$

$\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}$

$\mathbf{ r^2_iV_i = r^2_fV_f}$

$\mathbf{ (\dfrac{10}{2})^2\times 2.0 = r_f^2 \times 12.29}$

$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

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2 years ago

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

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$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

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so above is the relation between all given variable

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3 years ago

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

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$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

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$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

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3 years ago